How to find the limit of following sequences (natural logarithm)? [closed]
$lim_{xrightarrowinfty} ::ln x-x$
$lim_{xrightarrow0^+}::ln x-x$
limits logarithms
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26
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$lim_{xrightarrowinfty} ::ln x-x$
$lim_{xrightarrow0^+}::ln x-x$
limits logarithms
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
1
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22
add a comment |
$lim_{xrightarrowinfty} ::ln x-x$
$lim_{xrightarrow0^+}::ln x-x$
limits logarithms
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
$lim_{xrightarrowinfty} ::ln x-x$
$lim_{xrightarrow0^+}::ln x-x$
limits logarithms
limits logarithms
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
edited Dec 28 '18 at 14:24
Rakibul Islam Prince
1,035211
1,035211
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
asked Dec 28 '18 at 14:02
EmiliaEmilia
31
31
closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh Dec 28 '18 at 14:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, uniquesolution, KM101, José Carlos Santos, Siong Thye Goh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
1
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22
add a comment |
1
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
1
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22
1
1
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
1
1
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22
add a comment |
2 Answers
2
active
oldest
votes
You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
add a comment |
if $x>0$,we can expand as,
$$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
now,
$$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
$$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
$$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$
now,if
$$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
$$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
$$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
$$=-infty$$
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
add a comment |
You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
add a comment |
You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
You have that
$$log x-x=log(x)-log(e^x)=logleft(x/e^xright),$$
using continuity you have that in the first limit, $limlimits_{xtoinfty}log x-x=-infty$, and, $limlimits_{xto0^+}log x-x=-infty$.
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
answered Dec 28 '18 at 14:23
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
808110
808110
add a comment |
add a comment |
if $x>0$,we can expand as,
$$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
now,
$$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
$$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
$$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$
now,if
$$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
$$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
$$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
$$=-infty$$
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
add a comment |
if $x>0$,we can expand as,
$$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
now,
$$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
$$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
$$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$
now,if
$$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
$$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
$$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
$$=-infty$$
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
add a comment |
if $x>0$,we can expand as,
$$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
now,
$$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
$$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
$$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$
now,if
$$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
$$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
$$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
$$=-infty$$
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
if $x>0$,we can expand as,
$$ln x -x =2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
now,
$$lim_{x to infty} ln x -x=lim_{x to infty}2left[ frac{1-frac{1}{x}}{1+frac{1}{x}}+frac{1}{3}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^3+frac{1}{5}left(frac{1-frac{1}{x}}{1+frac{1}{x}}right)^5...right]-x$$
$$=2left[1+frac{1}{3}+frac{1}{5}+...right]-infty$$
$$=-infty$$$$text{we can ignore the series term,as it is way too smaller than $infty$}$$
now,if
$$lim_{x to 0^+} ln x -x=lim_{x to 0^+}2left[ frac{x-1}{x+1}+frac{1}{3}left(frac{x-1}{x+1}right)^3+frac{1}{5}left(frac{x-1}{x+1}right)^5...right]-x$$
$$=2left[-1-frac{1}{3}-frac{1}{5}+...right]-0^+$$
$$=-2left[1+frac{1}{3}+frac{1}{5}+...right]$$
$$=-infty$$
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
answered Dec 28 '18 at 14:47
Rakibul Islam PrinceRakibul Islam Prince
1,035211
1,035211
add a comment |
add a comment |
1
What have you tried? Have you plotted the graphs of the two functions to get some intuition? MathJax hint: if you put a backslash before lim, ln and other functions you get the right font and spacing, so ln(x) gives $ln(x)$ compared to ln(x) which gives $ln(x)$
– Ross Millikan
Dec 28 '18 at 14:10
How exactly would those limits be confusing?
– KM101
Dec 28 '18 at 14:10
1
Both limites are $$-infty$$
– Dr. Sonnhard Graubner
Dec 28 '18 at 14:22