# Calculating the limit: $\lim \limits_{x \to 0}$ $\frac{\ln(\frac{\sin x}{x})}{x^2}.$

How do I calculate $$\lim \limits_{x \to 0} \dfrac{\ln\left(\dfrac{\sin x}{x}\right)}{x^2}\text{?}$$

I thought about using L'Hôpital's rule, applying on "$\frac00$," but then I thought about $\frac{\sin x}{x}$ which is inside the $\ln$: it's not constant as $x$ goes to $0$, then I thought that maybe this what caused that my calculating of the limit wasn't true.

Can someone clarify how we calculate the limit here?

Note: I'd like to see a solution using L'Hôpital's rule.

• Do you know the limit of $\sin(x)/x$ as $x\to \infty$? Feb 13, 2015 at 21:22
• The limit of the full function is $-1/6$. Try l'Hospital's rule. Feb 13, 2015 at 21:23
• @mathmandan Shouldn't I check when x goes to 0 ? Feb 13, 2015 at 21:26
• I apologize. I should have said: "Do you know the limit of $\sin(x)/x$ as $x \to 0$? Feb 13, 2015 at 21:27
• @mathmandan yeah, its 1. and I applied L'Hopital's rule but didn't work! Feb 13, 2015 at 21:28

## 7 Answers

We want$$L = \lim_{x\to 0} \frac{\ln(\frac{\sin x}{x})}{x^2}$$ Since the top approaches $\ln(1) = 0$ and the bottom also approaches $0$, we may use L'Hopital:

$$L = \lim_{x\to 0}{\frac{(\frac{x}{\sin x})(\frac{x \cos x - \sin x}{x^2})}{2x}} = \lim_{x\to 0}\frac{x \cos x - \sin x}{2x^2\sin x}$$ Again the top and bottom both approach $0$, so again we may use L'Hopital.

$$L = \lim_{x \to 0} \frac{\cos x - x \sin x - \cos x}{4x \sin x + 2x^2 \cos x} = \lim_{x \to 0} \frac{-\sin x}{4 \sin x + 2x \cos x}$$

Again the top and bottom both approach $0$, so we may use L'Hopital for a third time:

$$L = \lim_{x \to 0} \frac{-\cos x}{4 \cos x + 2\cos x - 2x\sin x} = -\frac{1}{6}.$$

• Thank you!! this is what I've been trying to understand !! Feb 13, 2015 at 23:32

As $$\dfrac{\sin x}{x}=1-\dfrac{x^2}6+o(x^2)$$, we have: $$\frac{\ln\Bigl(\cfrac{\sin x}{x}\Bigr)}{x^2}=\frac{\ln\Bigl(1-\cfrac{x^2}6+o(x^2)\Bigr)}{x^2}=\frac{-\dfrac{x^2}6+o(x^2)}{x^2}=-\frac16+o(1),$$ which proves the limit is $$\,-\dfrac16$$.

• thank you. see the note that I edited. Feb 13, 2015 at 21:33
• @Bernard Oh no! I didn't see your answer dude! Feb 13, 2015 at 21:34
• Am I a ‘dude’, mylord? Feb 13, 2015 at 21:38
• Sorry, mylord, sir! Feb 13, 2015 at 21:38
• @Community: I'm sorry, but your edit is wrong. You seem to have confused the little-oh and the big-O notations. I use the former. Feb 19, 2017 at 16:39

Hint: Use finite expansion for $sin x$ in the neighberhood of zero! Will lead you to $ln(1-x^2/6)$ ! Using finite expansion again on $ln(-x^2/6+1)$ will lead to $-x^2/6$! You can evaluate the limit then which will then lead to $-1/6$ i guess! Sorry for my bad language.

Using taylor series you have

$$\color{#05f}{\frac{\sin x}{x}} = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n}}{(2n+1)!} = 1 - \frac{x^2}{6} + \frac{x^4}{120} + O(x^6)$$

and $$\ln \Bigg(\color{#05f}{\frac{\sin x}{x}}\Bigg) = - \frac{x^2}{6} - O(x^4)$$

$$\lim _{x\to 0 } \frac{- \frac{x^2}{6} - O(x^4)}{x^2}= \lim_{x\to 0} -\frac{1}{6} - O(x^2) = \color{#f05}{-\frac{1}{6}}$$

Edit:

By L'hospital you have

\begin{align}\require{cancel}\frac{d}{dx} \ln \Bigg(\frac{\sin x}{x}\Bigg) &= \frac{\color{#f05}{\cancel x}}{\sin x}\frac{x\cos x - \sin x}{x^{\color{#f05}{\cancel{2}}}} = \cot x - \frac{1}{x} \\&\implies \frac{d^2}{dx^2} \ln \Bigg(\frac{\sin x}{x}\Bigg) = \frac{1}{x^2} - \csc^2 x \color{#085}{\to -\frac{1}{3}} \end{align}

as $x \to 0$.

Then

$$\lim _{x\to 0 } \frac{\frac{1}{x^2} - \csc^2 x}{2}= {-\frac{1}{2}\frac{1}{3}} = \color{#f05}{-\frac{1}{6}}$$

• thank you. can u show a way of using L'Hôpital's rule? Feb 13, 2015 at 21:37
• There you have it. Feb 13, 2015 at 22:58
• but why the first one converges to $-1/3$ I just cant understand! Feb 13, 2015 at 23:04
• I've been trying to search for an email address of yours for mathematical consultation. (If u were able to help ofcourse) but couldn't find any. Feb 14, 2015 at 0:05
• Sure, It's on my profile. Feb 14, 2015 at 0:11

It is possible to compute \begin{equation*} \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{\ln \left( \frac{\sin x% }{x}\right) }{x^{2}} \end{equation*} only by making use of some basic limits: \begin{eqnarray*} \lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\ \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\ \lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1. \end{eqnarray*} No l'Hospital's rule nor Taylor series are required. Let \begin{eqnarray*} f(x) &=&\frac{1}{x^{2}}\ln \frac{\sin x}{x} \\ &=&\frac{1}{x^{2}}\ln \left( 1+[\frac{\sin x}{x}-1]\right) \\ &=&\frac{\left[ \frac{\sin x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+% \left[ \frac{\sin x}{x}-1\right] \right) }{\left[ \frac{\sin x}{x}-1\right] } \\ &=&\frac{\sin x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{% \sin x}{x}-1 \end{eqnarray*} since \begin{equation*} \lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\sin x}{x}-1)=1-1=0 \end{equation*} then \begin{equation*} \lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{% \ln (1+u)}{u}=1, \end{equation*} and therefore \begin{equation*} \lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\cdot \lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=-\frac{1}{6}\cdot 1=-\frac{1}{% 6}. \end{equation*}

Recall l'Hospital's rule: If the limit of $f(x)/g(x)$ for $x \rightarrow 0$ is undefined, and if the limit of $f'(x)/g'(x)$ is also undefined, then calculate $f''(x)/g''(x)$. In this case:

$\lim_{x\to 0} \, \frac{\partial ^2}{\partial x^2}\log \left(\frac{\sin (x)}{x}\right) = -1/3$

and

$\lim_{x\to 0} \, \frac{\partial ^2x^2}{\partial x^2} = 2$,

so the limit of the full term is $-1/6$.

• why undefined? $f(x)$/$g(x)$ its the limit of the type 0/0 Feb 13, 2015 at 21:47
• $0/0$ is undefined. If it isn't, please state the numerical value of it for me... what you would enter into your pocket calculator. Feb 13, 2015 at 21:49
• sorry. I misunderstood u. u used the rules twice because It wasn't defined. Feb 13, 2015 at 21:59
• @Stork: The limit is defined since it equals the real value $\frac{-1}{6}$. You're right to say $\frac{0}{0}$ is undefined, but the limit does not equal $\frac{0}{0}$. (hence his distinction of saying it is "of the type" $\frac{0}{0}$ rather than it "equals"). Feb 13, 2015 at 22:01
• I think what you meant to say is if the (limit of $f(x)$)/(limit of $g(x)$) is undefined. But that's not quite the same thing. So it was a little confusing. Feb 13, 2015 at 22:04

In these cases using a Taylor Series approach usually gets you to the solution: The denominator will be expanded as: $$\frac{-x^2}{6} + O(x^3)$$ Thus dividing by $x^2$ as $x\rightarrow 0$ leaves you with $\frac{-1}{6}$.