Why doesn't L'hopitals Rule work for $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$ [duplicate]
This question already has an answer here:
Why doesn't L'Hopital's rule work in this case?
4 answers
This is how I would evaluate $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$
$=limlimits_{x to infty} dfrac{x left( 1+ frac{sin x}{x} right)}{x left(1+ 2 cdot frac{ sin x}{x} right)}$
$= dfrac{1+0}{1+2 cdot 0} = 1$
But now applying L'hopitals Rule, I get
$limlimits_{x to infty} dfrac{1+ cos x}{1+ 2 cos x}$
Since $cos x $ just oscillates between $[-1,1]$ I think we can conclude the limit doesn't exist.
What is going on here?
limits
asked Dec 27 '18 at 13:44
William
1,170314
marked as duplicate by Zacky, amWhy, Pierre-Guy Plamondon, Eric Wofsey, Lord Shark the Unknown Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Why doesn't L'Hopital's rule work in this case?
4 answers
This is how I would evaluate $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$
$=limlimits_{x to infty} dfrac{x left( 1+ frac{sin x}{x} right)}{x left(1+ 2 cdot frac{ sin x}{x} right)}$
$= dfrac{1+0}{1+2 cdot 0} = 1$
But now applying L'hopitals Rule, I get
$limlimits_{x to infty} dfrac{1+ cos x}{1+ 2 cos x}$
Since $cos x $ just oscillates between $[-1,1]$ I think we can conclude the limit doesn't exist.
What is going on here?
limits
asked Dec 27 '18 at 13:44
William
1,170314
marked as duplicate by Zacky, amWhy, Pierre-Guy Plamondon, Eric Wofsey, Lord Shark the Unknown Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06
add a comment |
This question already has an answer here:
Why doesn't L'Hopital's rule work in this case?
4 answers
This is how I would evaluate $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$
$=limlimits_{x to infty} dfrac{x left( 1+ frac{sin x}{x} right)}{x left(1+ 2 cdot frac{ sin x}{x} right)}$
$= dfrac{1+0}{1+2 cdot 0} = 1$
But now applying L'hopitals Rule, I get
$limlimits_{x to infty} dfrac{1+ cos x}{1+ 2 cos x}$
Since $cos x $ just oscillates between $[-1,1]$ I think we can conclude the limit doesn't exist.
What is going on here?
limits
asked Dec 27 '18 at 13:44
William
1,170314
This question already has an answer here:
Why doesn't L'Hopital's rule work in this case?
4 answers
This is how I would evaluate $limlimits_{x to infty} dfrac{x+ sin x}{x+ 2 sin x}$
$=limlimits_{x to infty} dfrac{x left( 1+ frac{sin x}{x} right)}{x left(1+ 2 cdot frac{ sin x}{x} right)}$
$= dfrac{1+0}{1+2 cdot 0} = 1$
But now applying L'hopitals Rule, I get
$limlimits_{x to infty} dfrac{1+ cos x}{1+ 2 cos x}$
Since $cos x $ just oscillates between $[-1,1]$ I think we can conclude the limit doesn't exist.
What is going on here?
This question already has an answer here:
Why doesn't L'Hopital's rule work in this case?
4 answers
limits
limits
asked Dec 27 '18 at 13:44
William
1,170314
asked Dec 27 '18 at 13:44
William
1,170314
asked Dec 27 '18 at 13:44
William
1,170314
asked Dec 27 '18 at 13:44
William
1,170314
asked Dec 27 '18 at 13:44
William
1,170314
1,170314
marked as duplicate by Zacky, amWhy, Pierre-Guy Plamondon, Eric Wofsey, Lord Shark the Unknown Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Zacky, amWhy, Pierre-Guy Plamondon, Eric Wofsey, Lord Shark the Unknown Dec 28 '18 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06
add a comment |
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06
add a comment |
2 Answers
2
active
oldest
votes
L'Hospital's rule contains an assumption that $lim_{x to a} f'(x)/g'(x)$ exists, which is not true in this case.
answered Dec 27 '18 at 13:48
littleO
29.1k644108
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
add a comment |
Because your function doesn't satisfy the hypothesis. If you are studing the limit $xto c$, in order to apply the theorem the function $g=x+2sin x$ must be differentiable and $g'(x)ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+infty) where $g' ne 0$. But $g'(x)=0$ $forall x=-frac{pi}{4}+2kpi$, so it doesn't exists a set like that.
answered Dec 27 '18 at 14:02
ecrin
785
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
L'Hospital's rule contains an assumption that $lim_{x to a} f'(x)/g'(x)$ exists, which is not true in this case.
answered Dec 27 '18 at 13:48
littleO
29.1k644108
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
add a comment |
L'Hospital's rule contains an assumption that $lim_{x to a} f'(x)/g'(x)$ exists, which is not true in this case.
answered Dec 27 '18 at 13:48
littleO
29.1k644108
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
add a comment |
L'Hospital's rule contains an assumption that $lim_{x to a} f'(x)/g'(x)$ exists, which is not true in this case.
answered Dec 27 '18 at 13:48
littleO
29.1k644108
L'Hospital's rule contains an assumption that $lim_{x to a} f'(x)/g'(x)$ exists, which is not true in this case.
answered Dec 27 '18 at 13:48
littleO
29.1k644108
answered Dec 27 '18 at 13:48
littleO
29.1k644108
answered Dec 27 '18 at 13:48
littleO
29.1k644108
answered Dec 27 '18 at 13:48
littleO
29.1k644108
29.1k644108
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
add a comment |
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
3
3
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
Unfortunately, students often forget the hypotheses for a result when they try to apply it.
– GEdgar
Dec 27 '18 at 13:51
add a comment |
Because your function doesn't satisfy the hypothesis. If you are studing the limit $xto c$, in order to apply the theorem the function $g=x+2sin x$ must be differentiable and $g'(x)ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+infty) where $g' ne 0$. But $g'(x)=0$ $forall x=-frac{pi}{4}+2kpi$, so it doesn't exists a set like that.
answered Dec 27 '18 at 14:02
ecrin
785
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
add a comment |
Because your function doesn't satisfy the hypothesis. If you are studing the limit $xto c$, in order to apply the theorem the function $g=x+2sin x$ must be differentiable and $g'(x)ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+infty) where $g' ne 0$. But $g'(x)=0$ $forall x=-frac{pi}{4}+2kpi$, so it doesn't exists a set like that.
answered Dec 27 '18 at 14:02
ecrin
785
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
add a comment |
Because your function doesn't satisfy the hypothesis. If you are studing the limit $xto c$, in order to apply the theorem the function $g=x+2sin x$ must be differentiable and $g'(x)ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+infty) where $g' ne 0$. But $g'(x)=0$ $forall x=-frac{pi}{4}+2kpi$, so it doesn't exists a set like that.
answered Dec 27 '18 at 14:02
ecrin
785
Because your function doesn't satisfy the hypothesis. If you are studing the limit $xto c$, in order to apply the theorem the function $g=x+2sin x$ must be differentiable and $g'(x)ne 0$ in an open interval containing $c$, except in $c$. That means that you need a set (M,+infty) where $g' ne 0$. But $g'(x)=0$ $forall x=-frac{pi}{4}+2kpi$, so it doesn't exists a set like that.
answered Dec 27 '18 at 14:02
ecrin
785
answered Dec 27 '18 at 14:02
ecrin
785
answered Dec 27 '18 at 14:02
ecrin
785
answered Dec 27 '18 at 14:02
ecrin
785
785
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
add a comment |
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
The explanation is true and thus +1, yet the OP's question seems to point towards the fact that he forgot that L'Hospital's Rule applies in one direction only. Observe that if the denominator was $;3x+2sin x;$ then L'Hospital is appliable...yet it still wouldn't help as the limit of the quotient of the derivatives doesn't exist ...
– DonAntonio
Dec 27 '18 at 14:06
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
Thank you for the explanation!
– ecrin
Dec 27 '18 at 14:09
add a comment |
See also here: math.stackexchange.com/q/1342202/515527 . This is a question raised quite frequently.
– Zacky
Dec 27 '18 at 13:50
This is also explained on the wikipedia page (en.wikipedia.org/wiki/…) where there is a simpler example such as $(x + cos x)/x$.
– Ben
Dec 27 '18 at 13:51
See also math.stackexchange.com/questions/1710786/…
– Barry Cipra
Dec 27 '18 at 14:06