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$$\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^{\dfrac{1}{1 - \cos x}}$$


This limit is equal to,

$$\lim_{x \to 0} \exp\left({\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}\right).$$

I hope to solve this by focusing on the limit,

$$\lim_{x \to 0} {\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}.$$

Which is indeterminate of the form "$\frac{0}{0}$". However, repeated application of L'Hopital's Rule seems to lead to an endless spiral of trigonometric terms, but wolfram Alpha says the limit I am focusing on exists.

Is there another way to solve this?

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up vote 4 down vote accepted

It should be used. There may be some little mistakes in your computation.

$$\begin{align*} \lim_{x\rightarrow0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x} &=\lim_{x\rightarrow0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}\\ &=\lim_{x\rightarrow0}\frac{x\cos x-\sin x}{x\sin^{2}x}\\ &=\lim_{x\rightarrow0}\frac{-x\sin x}{\sin^{2}x+2x\sin x\cos x}\\ &=-\lim_{x\rightarrow0}\frac{1}{\frac{\sin x}{x}+2\cos x}\\ &=-\frac{1}{3} \end{align*}$$

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Thanks! I indeed made a mistake during the 3rd step of your calculation. – Legendre Oct 3 '12 at 12:29

You can try to expand $$ \log \frac{\sin x}{x} = -\frac{x^2}{6}-\frac{x^4}{180}+O(x^6) $$ and reduce to the limit of $$ \frac{-\frac{x^2}{6}-\frac{x^4}{180}+O(x^6)}{1-\cos x} = \frac{-\frac{x^2}{6}-\frac{x^4}{180}+O(x^6)}{\frac{x^2}{2}-\frac{x^4}{24}+O(x^6)}, $$ which is now easy to solve.

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No need to go to fourth order, though. – yohBS Oct 3 '12 at 12:04
Melius abundare quam deficere :-) When you expand a composition, it may be hard to guess what terms will cancel. – Siminore Oct 3 '12 at 12:06
+1 Thanks for the input. – Legendre Oct 3 '12 at 12:30
How do you get the first equality? – Jack Oct 12 '12 at 20:53
@Jack Expand $\frac{\sin x}{x}$ and plug into the expansion of $\log x$ around $x=1$. – Siminore Oct 13 '12 at 8:20

Sometimes there can be a lot of repetitions of l'Hopital needed. And here we even have some "nested" l'Hopital because of $\frac{\sin x}x$.

A nice way to avoid getting lost in l'Hopital bookkeeping is to work with power series expansions. This leads to $$\tag1\ln\left(\frac{\sin x}x\right)= -\frac16 x^2+O(x^4)$$ and $$1-\cos x= \frac12 x^2+O(x^4),$$ hence the quotient is just $-\frac13+O(x^2)$ and has limit $-\frac13$.

How does this work before one learns about Taylor series or if one does not even want to worry about convergence? Well, you may simply consider "$f(x)=a_0+a_1x+\ldots +a_{n-1}x^{n-1}+O(x^n)$" simply as a friendly encoding of "$f$ is at least $n-1$ times continuously differentiable at $x=0$ and $f^{(k)}(0)=k!a_k$ for $0\le k<n$". It follow sthat the most obvious rules apply for combining such results. Especially, $\sin x=x-\frac16x^3+O(x^5)$ implies $\frac{\sin x}x=1-\frac16x^2+O(x^4)$. Together with $\ln(1+x)=x+O(x^2)$ this produces $(1)$ above. $(2)$ follows simply from $\cos x=1-\frac12x^2+O(x^4)$. Cancellation of $x^2$ and dropping the $O$ terms then is equivalent to applying l*Hopital to times.

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+1 very nice explanation. – Legendre Oct 3 '12 at 12:30

Use $\ln(\frac{\sin x}{x})=\ln(\sin x)-\ln x$

Then, $$\frac{d(\ln(\frac{\sin x}{x}))}{dx}=\cot x-\frac{1}{x}=\frac{x\cos x-\sin x}{x\sin x}$$

and $$\frac{d(1-\cos x)}{dx}=\sin x$$ which gives $$\lim_{x \to 0} {(\frac{1}{1 - \cos x})\ln (\frac{\sin x}{x})}=\lim_{x \to 0} \frac{(x\cos x-\sin x)}{x\sin^2 x}$$ which again applying L'Hopital's rule gives $$\lim_{x \to 0} \frac{(x\cos x-\sin x)}{x\sin^2 x}=\lim_{x \to 0} \frac{-x\sin x}{2x\sin x\cos x+\sin^2 x}=\lim_{x \to 0} \frac{-1}{2\cos x+\frac{\sin x}{x}}=-\frac{1}{3}$$ since $\lim_{x\to 0}\frac{\sin x}{x}=1$

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+1, thanks! Question: Doesn't $log(xy) = log(x) + log(y)$ only works if $x,y$ are positive? Since $sin x$ can be negative, should we be doing that? – Legendre Oct 3 '12 at 12:28
@Legendre That's a very good observation, and let me share an fix that might patch this problem. Sufficently close to 0, (on the interval $[-\pi/2,\pi/2]$ will do) $\dfrac{\sin x}{x}=\dfrac{\sin |x|}{|x|}$, and so the rule that $\dfrac{d}{dx}\ln(|x|)=\dfrac{1}{x}$ might be useful. I haven't checked to see if the absolute values in the sine cause any problem, but I'm imagining it's a two case check at worst. The original proposition is interesting, at least :) – rschwieb Oct 3 '12 at 12:42

One can also simplify as follows:

$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(\dfrac{\sin x}{x}\right)\\ &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(1 + \dfrac{\sin x}{x} - 1\right)\\ &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\left(\dfrac{\sin x}{x} - 1\right)}\\ &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\cdot 1\\ &= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot \dfrac{\sin x - x}{x^{3}}\\ &= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\sin x - x}{x^{3}}\\ &= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\cos x - 1}{3x^{2}}\text{ (by L'Hospital's Rule)}\\ &= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\dfrac{\cos x - 1}{3x^{2}}\\ &= -\frac{1}{3}\end{aligned}$

This solves the problem with minimal use of L'Hospital Rule. We have made use of the standard limit $\lim_{y \to 0}\dfrac{\log (1 + y)}{y} = 1$ where $y = \dfrac{\sin x}{x} - 1$ which tends to $0$ as $x \to 0$.

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