Find $lim _{xto infty }x^{frac{1}{x}}$ without L'Hôpital's Rule
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$$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.
I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).
Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?
limits limits-without-lhopital
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
$endgroup$
add a comment |
$begingroup$
$$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.
I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).
Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?
limits limits-without-lhopital
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
$endgroup$
add a comment |
$begingroup$
$$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.
I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).
Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?
limits limits-without-lhopital
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
$endgroup$
$$lim _{xto infty }x^{frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.
I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/infty,$ basic logs).
Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?
limits limits-without-lhopital
limits limits-without-lhopital
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
edited Jul 27 '16 at 19:16
Max Li
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
asked Jul 27 '16 at 19:06
Max LiMax Li
860619
860619
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write
$$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$
Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that
$$lim_{xtoinfty}frac{log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
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nice answer,without l'hospital rule ,+1
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– haqnatural
Jul 27 '16 at 19:28
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What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
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@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
add a comment |
$begingroup$
firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
so we can apply L'hospital's rule
$$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
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Can you do this step-by-step providing sources or explanations for the properties you have used?
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– Max Li
Jul 27 '16 at 19:10
1
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@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
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– Simply Beautiful Art
Jul 27 '16 at 19:12
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Is there a way without L'Hôpital's Rule?
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– Max Li
Jul 27 '16 at 19:17
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yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
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– haqnatural
Jul 27 '16 at 19:20
1
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@Bernard Indeed so...but almost .
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– DonAntonio
Jul 27 '16 at 19:25
|
show 2 more comments
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$x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.
You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$
edited Jan 16 at 16:42
user376343
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
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Can you do this step-by-step providing sources or explanations for the properties you have used?
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– Max Li
Jul 27 '16 at 19:10
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Shouldn't it be $ln(x)$ not $log(x)$?
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– Max Li
Jul 27 '16 at 19:18
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@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
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– Zain Patel
Jul 27 '16 at 19:19
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@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
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– Simply Beautiful Art
Jul 27 '16 at 19:32
add a comment |
$begingroup$
If you consider this
for integer $n$,
you want to show that
$lim_{n to infty} n^{1/n}
= 1
$.
Here's a bit of magic:
By Bernoulli's inequality,
$(1+frac1{sqrt{n}})^n
ge 1+nfrac1{sqrt{n}}
ge 1+sqrt{n}
gt sqrt{n}
= n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$((1+frac1{sqrt{n}})^n)^{2/n}
> (n^{1/2})^{2/n}
$
or
$(1+frac1{sqrt{n}})^2
> n^{1/n}
$
or
$n^{1/n}
< (1+frac1{sqrt{n}})^2
=1+2frac1{sqrt{n}}+frac1{n}
<1+3frac1{sqrt{n}}
$.
Since
$n^{1/n} > 1$
and
$lim_{n to infty}frac1{sqrt{n}}
=0
$,
$lim_{n to infty} n^{1/n}
=
1
$.
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
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add a comment |
$begingroup$
As to avoiding the use of L'Hospital's rule:
$$ln(x)=O(sqrt x)$$
$$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$
$$=lim_{xtoinfty}frac1{sqrt x}=0$$
And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
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add a comment |
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Let $x=1+frac {1}{y}$ Where $y rightarrow 0$
Then,
$lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
= lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$
$=lim _{y to 0}{1^{frac {1}{1+y}}}=1$
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
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The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
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– DonAntonio
Jul 27 '16 at 19:58
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Yes, the step before the last step should be $={1^{frac {1}{1}}}$
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– Pentapolis
Jul 27 '16 at 20:50
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If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
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– Physicist137
Aug 9 '16 at 18:39
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$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
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– Pentapolis
Aug 9 '16 at 22:09
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Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
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– Physicist137
Aug 10 '16 at 1:05
|
show 1 more comment
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write
$$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$
Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that
$$lim_{xtoinfty}frac{log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
add a comment |
$begingroup$
If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write
$$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$
Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that
$$lim_{xtoinfty}frac{log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
add a comment |
$begingroup$
If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write
$$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$
Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that
$$lim_{xtoinfty}frac{log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
$endgroup$
If you know that for any $;epsilon>0;$ we have that $;log x<x^epsilon;$ for any $;0<xinBbb R;$ big enough , you then don't need l'Hospital since we can write
$$frac{log x}x<frac{x^epsilon}x=frac1{x^{1-epsilon}}xrightarrow[xtoinfty]{}0;,;;text{for say};;0<epsilon<1$$
Using that $;log x>log1=0;$ for $;x>1;$, use the squeeze theorem to deduce that
$$lim_{xtoinfty}frac{log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{frac1xlog x}implieslim_{xtoinfty}x^{1/x}=e^{limlimits_{xtoinfty}frac1xlog x}=e^0=1$$
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
edited Jul 27 '16 at 19:29
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
answered Jul 27 '16 at 19:24
DonAntonioDonAntonio
180k1494233
180k1494233
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
add a comment |
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
nice answer,without l'hospital rule ,+1
$endgroup$
– haqnatural
Jul 27 '16 at 19:28
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
What is an elementary way to show the first claim (that $log x = O(x^epsilon)$)?
$endgroup$
– angryavian
Jul 27 '16 at 19:48
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@angryavian I'm not sure. I think I can prove it by means of differentials but I'm not sure if this makes the cut of "elementary".
$endgroup$
– DonAntonio
Jul 27 '16 at 19:56
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
$begingroup$
@DonAntonio I guess I meant (ideally) without calculus, or at at least something at the same level as the rest of your answer. Is there something along the lines of this answer?
$endgroup$
– angryavian
Jul 27 '16 at 20:22
add a comment |
$begingroup$
firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
so we can apply L'hospital's rule
$$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
$endgroup$
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
1
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
1
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
|
show 2 more comments
$begingroup$
firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
so we can apply L'hospital's rule
$$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
$endgroup$
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
1
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
1
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
|
show 2 more comments
$begingroup$
firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
so we can apply L'hospital's rule
$$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
$endgroup$
firstly,write it as $x^{ frac { 1 }{ x } }={ e }^{ frac { 1 }{ x } ln { x } }$
now we have such limit $$lim _{ xto infty } e^{ frac { ln { x } }{ x } }=lim _{ xrightarrow infty }{ { e }^{ frac { fleft( x right) rightarrow infty }{ gleft( x right) rightarrow infty } } } ,quad wherequad fleft( x right) =ln { x } ,gleft( x right) =x $$
so we can apply L'hospital's rule
$$lim _{ xto infty } x^{ frac { 1 }{ x } }=lim _{ xto infty } e^{ frac { 1 }{ x } ln { x } }=lim _{ xto infty } e^{ frac { 1 }{ x } }=1$$
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
edited Jul 27 '16 at 19:14
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
answered Jul 27 '16 at 19:08
haqnaturalhaqnatural
20.8k72457
20.8k72457
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
1
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
1
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
|
show 2 more comments
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
1
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
1
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
1
1
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
@MaxLi The first two are algaebraically equal. The second and third involve L'Hopital's Rule. And the last is evaluating the limit.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:12
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
Is there a way without L'Hôpital's Rule?
$endgroup$
– Max Li
Jul 27 '16 at 19:17
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
$begingroup$
yes,you can prove that $x>ln { x } ,x>1$ so that x "runs" to infinity more quickly than $lnx$
$endgroup$
– haqnatural
Jul 27 '16 at 19:20
1
1
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
$begingroup$
@Bernard Indeed so...but almost .
$endgroup$
– DonAntonio
Jul 27 '16 at 19:25
|
show 2 more comments
$begingroup$
$x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.
You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$
edited Jan 16 at 16:42
user376343
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
add a comment |
$begingroup$
$x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.
You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$
edited Jan 16 at 16:42
user376343
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
add a comment |
$begingroup$
$x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.
You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$
edited Jan 16 at 16:42
user376343
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$x^{1/x}=e^{ln(x)/x}$ so the limit is $limlimits_{xrightarrow +infty}e^{ln(x)/x}$.
You know that $limlimits_{+infty}ln(x)/x=0$ this implies that $limlimits_{xrightarrow +infty}e^{ln(x)/x}=1.$
edited Jan 16 at 16:42
user376343
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
edited Jan 16 at 16:42
user376343
3,9584829
edited Jan 16 at 16:42
user376343
3,9584829
edited Jan 16 at 16:42
user376343
3,9584829
3,9584829
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
answered Jul 27 '16 at 19:08
Tsemo AristideTsemo Aristide
60.2k11446
60.2k11446
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
add a comment |
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Can you do this step-by-step providing sources or explanations for the properties you have used?
$endgroup$
– Max Li
Jul 27 '16 at 19:10
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
Shouldn't it be $ln(x)$ not $log(x)$?
$endgroup$
– Max Li
Jul 27 '16 at 19:18
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi $log = ln$ for most purposes. Also, passing the limit to the exponent is justified by continuity of $exp$.
$endgroup$
– Zain Patel
Jul 27 '16 at 19:19
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
$begingroup$
@MaxLi Under the right context (computer science) $log$ can imply $log_2$. Usually, $log=log_{10}$, and if we are doing limits/calculus, $log=log_e=ln$.
$endgroup$
– Simply Beautiful Art
Jul 27 '16 at 19:32
add a comment |
$begingroup$
If you consider this
for integer $n$,
you want to show that
$lim_{n to infty} n^{1/n}
= 1
$.
Here's a bit of magic:
By Bernoulli's inequality,
$(1+frac1{sqrt{n}})^n
ge 1+nfrac1{sqrt{n}}
ge 1+sqrt{n}
gt sqrt{n}
= n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$((1+frac1{sqrt{n}})^n)^{2/n}
> (n^{1/2})^{2/n}
$
or
$(1+frac1{sqrt{n}})^2
> n^{1/n}
$
or
$n^{1/n}
< (1+frac1{sqrt{n}})^2
=1+2frac1{sqrt{n}}+frac1{n}
<1+3frac1{sqrt{n}}
$.
Since
$n^{1/n} > 1$
and
$lim_{n to infty}frac1{sqrt{n}}
=0
$,
$lim_{n to infty} n^{1/n}
=
1
$.
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
$endgroup$
add a comment |
$begingroup$
If you consider this
for integer $n$,
you want to show that
$lim_{n to infty} n^{1/n}
= 1
$.
Here's a bit of magic:
By Bernoulli's inequality,
$(1+frac1{sqrt{n}})^n
ge 1+nfrac1{sqrt{n}}
ge 1+sqrt{n}
gt sqrt{n}
= n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$((1+frac1{sqrt{n}})^n)^{2/n}
> (n^{1/2})^{2/n}
$
or
$(1+frac1{sqrt{n}})^2
> n^{1/n}
$
or
$n^{1/n}
< (1+frac1{sqrt{n}})^2
=1+2frac1{sqrt{n}}+frac1{n}
<1+3frac1{sqrt{n}}
$.
Since
$n^{1/n} > 1$
and
$lim_{n to infty}frac1{sqrt{n}}
=0
$,
$lim_{n to infty} n^{1/n}
=
1
$.
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
$endgroup$
add a comment |
$begingroup$
If you consider this
for integer $n$,
you want to show that
$lim_{n to infty} n^{1/n}
= 1
$.
Here's a bit of magic:
By Bernoulli's inequality,
$(1+frac1{sqrt{n}})^n
ge 1+nfrac1{sqrt{n}}
ge 1+sqrt{n}
gt sqrt{n}
= n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$((1+frac1{sqrt{n}})^n)^{2/n}
> (n^{1/2})^{2/n}
$
or
$(1+frac1{sqrt{n}})^2
> n^{1/n}
$
or
$n^{1/n}
< (1+frac1{sqrt{n}})^2
=1+2frac1{sqrt{n}}+frac1{n}
<1+3frac1{sqrt{n}}
$.
Since
$n^{1/n} > 1$
and
$lim_{n to infty}frac1{sqrt{n}}
=0
$,
$lim_{n to infty} n^{1/n}
=
1
$.
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
$endgroup$
If you consider this
for integer $n$,
you want to show that
$lim_{n to infty} n^{1/n}
= 1
$.
Here's a bit of magic:
By Bernoulli's inequality,
$(1+frac1{sqrt{n}})^n
ge 1+nfrac1{sqrt{n}}
ge 1+sqrt{n}
gt sqrt{n}
= n^{1/2}
$.
Raising both sides
to the $2/n$ power,
$((1+frac1{sqrt{n}})^n)^{2/n}
> (n^{1/2})^{2/n}
$
or
$(1+frac1{sqrt{n}})^2
> n^{1/n}
$
or
$n^{1/n}
< (1+frac1{sqrt{n}})^2
=1+2frac1{sqrt{n}}+frac1{n}
<1+3frac1{sqrt{n}}
$.
Since
$n^{1/n} > 1$
and
$lim_{n to infty}frac1{sqrt{n}}
=0
$,
$lim_{n to infty} n^{1/n}
=
1
$.
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
answered Jul 27 '16 at 20:32
marty cohenmarty cohen
75k549130
75k549130
add a comment |
add a comment |
$begingroup$
As to avoiding the use of L'Hospital's rule:
$$ln(x)=O(sqrt x)$$
$$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$
$$=lim_{xtoinfty}frac1{sqrt x}=0$$
And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
$endgroup$
add a comment |
$begingroup$
As to avoiding the use of L'Hospital's rule:
$$ln(x)=O(sqrt x)$$
$$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$
$$=lim_{xtoinfty}frac1{sqrt x}=0$$
And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
$endgroup$
add a comment |
$begingroup$
As to avoiding the use of L'Hospital's rule:
$$ln(x)=O(sqrt x)$$
$$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$
$$=lim_{xtoinfty}frac1{sqrt x}=0$$
And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
$endgroup$
As to avoiding the use of L'Hospital's rule:
$$ln(x)=O(sqrt x)$$
$$implieslim_{xtoinfty}frac{ln(x)}x=lim_{xtoinfty}frac{sqrt x}x$$
$$=lim_{xtoinfty}frac1{sqrt x}=0$$
And so we have $lim x^{1/x}=e^{ln(x)/x}=e^0=1$
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
answered Jul 27 '16 at 19:35
Simply Beautiful ArtSimply Beautiful Art
50.9k580186
50.9k580186
add a comment |
add a comment |
$begingroup$
Let $x=1+frac {1}{y}$ Where $y rightarrow 0$
Then,
$lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
= lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$
$=lim _{y to 0}{1^{frac {1}{1+y}}}=1$
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
$endgroup$
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
|
show 1 more comment
$begingroup$
Let $x=1+frac {1}{y}$ Where $y rightarrow 0$
Then,
$lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
= lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$
$=lim _{y to 0}{1^{frac {1}{1+y}}}=1$
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
$endgroup$
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
|
show 1 more comment
$begingroup$
Let $x=1+frac {1}{y}$ Where $y rightarrow 0$
Then,
$lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
= lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$
$=lim _{y to 0}{1^{frac {1}{1+y}}}=1$
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
$endgroup$
Let $x=1+frac {1}{y}$ Where $y rightarrow 0$
Then,
$lim _{xto infty }x^{frac{1}{x}} =lim _{yto 0 }{(1+frac {1}{y})}^{frac {y}{1+y}}
= lim _{yto 0 }{(1+frac {1}{y})}^{yfrac {1}{1+y}}$
$=lim _{y to 0}{1^{frac {1}{1+y}}}=1$
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
answered Jul 27 '16 at 19:49
PentapolisPentapolis
36419
36419
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
|
show 1 more comment
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
The step from the third line to the last one is, to say the least, very questionable: how could you justify to take the limit of only a part of the expression in the right extreme of the line one before the last, but to leave other part still with $;y;$ ?
$endgroup$
– DonAntonio
Jul 27 '16 at 19:58
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
Yes, the step before the last step should be $={1^{frac {1}{1}}}$
$endgroup$
– Pentapolis
Jul 27 '16 at 20:50
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
If $yto 0$, then $1/ytoinfty$. Then in the end you have: $1+infty$, which is evidently not $1$.
$endgroup$
– Physicist137
Aug 9 '16 at 18:39
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
$lim _{yto 0 }{(1+frac {1}{y})}^y=1$ @Physicist137 this limit is well known which is equal to 1.
$endgroup$
– Pentapolis
Aug 9 '16 at 22:09
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
$begingroup$
Oh! Obviously. You are right. My mistake. Sorry... =). And +1. =).
$endgroup$
– Physicist137
Aug 10 '16 at 1:05
|
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