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I was discussing L'Hôpital's Rule with a Calculus I student earlier today. I mentioned that if the limit obtained by differentiating the numerator and denominator doesn't exist, then L'Hôpital's Rule tells us nothing about the original limit.

A clear example of this is, $$\lim\limits_{ x \to \infty }{ \frac { x+\sin { x } }{ x } } =1.$$ However, L'Hôpital's Rule gives $$\lim\limits_{ x\rightarrow \infty }{ \frac { x+\sin { x } }{ x } } =\lim \limits_{ x\rightarrow \infty }{ \frac { 1+\cos { x } }{ 1 } } =\lim\limits_{ x\rightarrow \infty }{ \left( 1+\cos { x } \right) }, $$ which diverges by oscillation.

I couldn't come up with an example that shows that if the limit from LH is infinite, then the original limit may be finite. This begs these two questions,

  1. Is it true that an infinite result from L'Hôpital's Rule does not imply an infinite limit?
  2. Is there a simple example where the LH is infinite, but the limit is actually finite?
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The links you provided address the problem of oscillation, but not an infinite limit. Am I mistaken in thinking that a infinite limit from LH doesn't imply an infinite original limit? – Gamma Function Apr 16 '13 at 1:25
If $\lim_{x \to 0} \frac f g= \frac 0 0$ and $\lim_{x \to 0} \frac{f'}{g'}=\infty$ then LH holds and $\lim_{x \to 0} \frac f g=\infty$. I don't think any example exists. LH is okay when the limit is infinite, just not when the limit doesn't exist. – spitespike Apr 16 '13 at 1:32
@spitespike I didn't realize that, I was thinking that the limit had to exist and be finite. Thanks for the clarification. – Gamma Function Apr 16 '13 at 1:36
up vote 13 down vote accepted

The statement of l'Hopital's rule found in Rudin's Principles of Mathematical Analysis (page 109) is:

$5.13$ $\,\,$ Theorem $\,\,\,\,\,\,\,$ Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty\leq a < b\leq +\infty$. Suppose $$ \frac{f'(x)}{g'(x)}\to A\,\textrm{ as }\,x\to a. $$ If $f(x)\to 0$ and $g(x)\to 0$ as $x\to a$, or if $g(x)\to +\infty$ as $x\to a$, then $$ \frac{f(x)}{g(x)}\to A\,\textrm{ as }\,x\to a. $$ The analogous statement is of course also true if $x\to b$, or if $g(x)\to-\infty$ [...]. Let us note that we now use the limit concept in the extended sense of Definition $4.33$.

It seems that what $A$ is supposed to be is a little ambiguous. Real? Possibly infinite? However, we can resolve this enigma with a quick look at definition $4.33$:

$4.33$ $\,\,$ Definition $\,\,\,\,\,\,\,$ Let $f$ be a real function defined on $E\subset \Bbb{R}$. We say that $$ f(t)\to A\,\textrm{ as }\,t\to x, $$ where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V\cap E$ is not empty, and such that $f(t)\in U$ for all $t\in V\cap E$, $t\neq x$.

So indeed, $A$ is allowed to be infinite (it's in the extended real number system)! We also see that Rudin treats the cases of $A = \pm\infty$ in the proof of l'Hopital's rule, so if the limit of the quotient of derivatives is infinite, the original limit must have been also. That is, there is no example (let alone a simple one) where the limit of the quotient of derivatives is infinite, but the original limit is finite.

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I assumed that the definition required the limit to exist and be finite. Thanks for the clarification! – Gamma Function Apr 16 '13 at 1:34
The extended versions of l'Hospital's rule are discussed in many prior posts here, e.g. this one, which includes some references to expository articles. – Math Gems Apr 16 '13 at 1:46

The problem can be solved without applying L'Hospital rule $$\lim_{x\to\infty}\frac{x+\sin x}x =\lim_{x\to\infty} 1 + \frac{\sin x}x$$

since for any $x$, $\sin x$ will be between $-1$ and $1$, where as $1/x$ value tends to $0$

so the answer to this question is $1$. without applying L'Hospital's rule (L's rule is applied when the limit is indeterminate and in this case, it is not)

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Remember, L'Hôpital's Rule states: Let $c$ finite or infinite. Suppose $\lim_{x\to c}{f(x)} = \lim_{x\to c}g(x) = 0$ (or $\lim_{x\to c}{|f(x)|} = \lim_{x\to c}{|g(x)|} = \infty$) and $\lim_{x\to c}{\frac{f(x)}{g(x)}}=L$. If $g'(x)\neq 0$ near $c$ and $\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L$, then $$ \lim_{x\to c}{\frac{f(x)}{g(x)}}=\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L. $$ In your example, this condition $\lim_{x\to c}{\frac{f'(x)}{g'(x)}} = L$ is violated. So you can not use LH to the limit.

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I didn't down-vote. But his concern is Where LH fails?. – Inceptio Apr 16 '13 at 1:41
I already said that $\lim_{x\to c}\frac{f'(x)}{g'(x)}=L$ is violated. – xpaul Apr 16 '13 at 2:02

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