Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to calculate the following limit:

$$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{x}}}$$

What I did is writing it as:

$${e^{\frac{1}{x}\ln \left( {\frac{{\sin x}}{x}} \right)}}$$

Therefore, we need to calculate:

$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x}$$

Now, we can apply L'Hopital rule, Which I did:
$$\Rightarrow cot(x) - {1 \over x}$$

But in order to reach the final limit two more application of LHR are needed. Is there a better way?

share|cite|improve this question

4 Answers 4

up vote 8 down vote accepted

Note that as $x\to 0^+$ you have $$\frac{\ln\frac{\sin x}x}x\sim \frac{\frac{\sin x}x-1}x\sim \frac{\sin x-x}{x^2}\sim \frac{-x^3}{6x^2}\to 0$$.

share|cite|improve this answer
How can I explain those steps? I understand what you did, but how can I write it down formally? certainly, those tildas won't do the work :) – SuperStamp Jan 11 '14 at 23:29
The tilda means asmiphtotic (see, so my writing is formally complete and correct. In particular, I applied $\ln(1+x)\sim x$ and $\sin(x)-x\sim -\frac{x^3}6$ as $x\to 0$. – Fabio Lucchini Jan 11 '14 at 23:31
hmm..OK, didn't know that. Thanks! – SuperStamp Jan 11 '14 at 23:32
Would you add verbal explanation to this in a proof? – SuperStamp Jan 11 '14 at 23:32
I used an "asymptotic" version of Taylor-Mac Laurin formula for $\sin$ and $\ln$. – Fabio Lucchini Jan 11 '14 at 23:38

Elementary proof using well known limits and inequalities: $$\frac{{\ln \left( {\frac{{\sin x}}{x}} \right)}}{x} = \frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\frac{{\sin x}}{x}-1 }{x} =\frac{{\ln \left( \left( \frac{{\sin x}}{x}-1 \right) + 1 \right)}}{\frac{{\sin x}}{x}-1 } \cdot \frac{\sin x-x }{x^2}$$ but for $x >0$ we have: $x-\frac{x^3}{6} \le \sin x \le x \Rightarrow -\frac{x^3}{6} = x-\frac{x^3}{6}-x\le\sin x -x \le 0 \Rightarrow -\frac{x^2}{6} \le\frac{\sin x-x }{x^2} \le 0$ the rest is squeeze theorem.

share|cite|improve this answer
nice one, I don't think I could come up with that. – SuperStamp Jan 11 '14 at 23:46

According to Bernoulli's Inequality, for $0\le x\le1$ $$ \left(\frac{\sin(x)}{x}\right)^{1/x}=\left(1-\frac{x-\sin(x)}{x}\right)^{1/x}\ge1-\frac1x\frac{x-\sin(x)}{x}\tag{1} $$ This answer provides an elementary proof of Bernoulli's Inequality for rational exponents.

In this answer, it is shown geometrically that for $0\lt x\lt\frac\pi2$, we have $\sin(x)\le x\le\tan(x)$. Therefore, $$ \begin{align} 0\le\frac{x-\sin(x)}{x^2}&\le\frac{\tan(x)-\sin(x)}{\sin^2(x)}\\[4pt] &=\frac{\tan(x)-\sin(x)}{\sin(x)\tan(x)}\sec(x)\\[4pt] &=\frac{1-\cos(x)}{\sin(x)}\sec(x)\\[4pt] &=\frac{\sin(x)}{1+\cos(x)}\sec(x)\\[4pt] &\to\frac02\cdot1\\[9pt] &=0\tag{2} \end{align} $$ Thus, by the Squeeze Theorem, $$ \lim_{x\to0}\frac{x-\sin(x)}{x^2}=0\tag{3} $$ and therefore, $$ 1\ge\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{1/x}\ge1-0\tag{4} $$

share|cite|improve this answer
Thank you sir, nice approach! – SuperStamp Jan 11 '14 at 23:53
@SuperStamp: I've just removed L'Hospital :-) – robjohn Jan 12 '14 at 0:16
Why did you remove LHR? – SuperStamp Jan 12 '14 at 0:18
To make the limit more elementary. I thought it would be nice to use only pre-calculus concepts. – robjohn Jan 12 '14 at 0:21
Cool. very appreciated! :) – SuperStamp Jan 12 '14 at 0:22

Here, as often, Taylor series are much more efficient than L'Hopital's rule.

$$\begin{eqnarray}\dfrac{\sin(x)}x \ &=&\ \color{#c00}1\color{#0a0}{\,-\,\dfrac{x^2}6+O(x^4)}\!\!\!\!\!\!\!\\&=&\ \color{#c00}1\, +\,\color{#0a0}z\\ \Rightarrow\ \ x^{-1}\log\left(\dfrac{\sin(x)}x\right) &=&\ x^{-1}\log(\color{#c00}1+\color{#0a0}z)\ &=&\ x^{-1}\left(\ \ \ \color{#0a0}z\ \ \ +\ \ O(z^2)\right)\\ && &=&\ x^{-1}\left(-\color{#0a0}{\dfrac{x^2}6+O(x^4)} \right)\\ && &=&\ -\dfrac{x}6\, +\ O(x^3)\\ && &&\!\!\!\!\!\!\!{\rm which}\overset{\phantom{I}} \to \ 0 \ \ \, {\rm as}\, \ \ x\to0^+\\ \end{eqnarray}\qquad\quad\qquad\qquad$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.