Help Evaluating $\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$

Question

Does anyone know how to evaluate the following limit?

$\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$

The answer is 1 , but I want to see a step by step solution if possible.

Maybe you could let $a$ equal the limit and take the natural logarithm of the resulting equality. I imagine L'Hopital might also be of some use thereafter. — Khallil, Aug 24 '14 at 19:57
Related: http://math.stackexchange.com/questions/632527/finding-the-limit-displaystyle-lim-x-to-0-left-frac-sin-xx-right1-x This exact one did come up recently. — colormegone, Aug 24 '14 at 21:07

score 5 · Answer 1 · answered Aug 24 '14 at 20:01

Define

$$F(x)=\frac{1}{x}\log{\left(\frac{\sin{x}}{{x}}\right)}$$

It is now suffice to prove that $$\lim_{x\to0}F(x)=0$$

Since near $x=0$, we have $$\log{\left(\frac{\sin{x}}{{x}}\right)}=-\frac{x^2}{6}+O(x^4)$$

So

$$\lim_{x\to0}F(x)=\lim_{x\to0}\left(-\frac{x}{6}\right)=0$$

Mhenni Benghorbal · Answer 2 · 2014-08-24T20:04:29.197

2

Hint: First write the expression as

$$e^{\frac{1}{x}\ln(\frac{\sin(x)}{x})} = e^{\frac{1}{x}(-\frac{x^2}{6}+O(x^4))}\longrightarrow_{x\to 0} ...\,. $$

using the Taylor series of $\ln(\frac{\sin(x)}{x})$.

Note:

$$ \ln\left(\frac{\sin(x)}{x}\right) = -\frac{x^2}{6}+O(x^4). $$

edited Aug 24 '14 at 20:04

answered Aug 24 '14 at 19:53

Mhenni Benghorbal

47,431

user84413 · Answer 3 · 2014-08-24T20:24:04.450

2

Let $\displaystyle y=(\frac{\sin x}{x})^{\frac{1}{x}}$. Then $\displaystyle\lim_{x\to 0} \ln y=\lim_{x\to 0}\ln \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to 0}\frac{\ln\sin x-\ln x}{x}=^{(LH)}\lim_{x\to 0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{1}$

$\displaystyle=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\sin x+\cos x-\cos x}{x\cos x+\sin x}=^{(LH)}\lim_{x\to 0}\frac{-x\cos x-\sin x}{-x\sin x+\cos x+\cos x}=\frac{0}{2}=0,$

so $\displaystyle\lim_{x\to 0}y=e^0=1$.

edited Aug 24 '14 at 20:24

answered Aug 24 '14 at 20:15

user84413

27,211

It might be more helpful to the OP if you specified where you applied L'Hopital's rule! Nonetheless, nice solution! – Khallil Aug 24 '14 at 20:16
@Khallil That is a good suggestion; I will see if I can do this. – user84413 Aug 24 '14 at 20:20
You could use \overset{\text{LH}}= (which is $\overset{\text{LH}}=$) to denote LH above the equal sign! – Khallil Aug 24 '14 at 20:28
@Khallil Thank you; I didn't know the right way to do this. – user84413 Aug 24 '14 at 20:30

score 0 · Accepted Answer · answered Aug 24 '14 at 20:21

First write limit in the other form:

$$\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}=\lim_{x \to 0}e^{\ln\frac{\sin{x}}{{x}}}=\lim_{x \to 0}e^{\frac{1}{x}\ln{\frac{\sin x}{x}}}=e^{\lim_{x \to 0}\frac{1}{x}\ln{\frac{sin}{x}}}$$

Now evaluate limit $\displaystyle \lim_{x \to 0}\frac{1}{x}\ln{(\frac{sin}{x})^{\frac{1}{x}}}$. You can use l'Hospital rule.

$$\lim_{x \to 0}\frac{1}{x}\ln{\frac{sin}{x}}=\lim_{x \to 0}\frac{(\ln\frac{sin}{x})'}{x'}=\lim_{x \to 0} (\frac{\sin x}{x})'\cdot \frac{x}{\sin x} =\lim_{x \to 0}\frac{x}{\sin x}\cdot \lim_{x \to 0}(\frac{\sin x}{x})'$$

Because $\displaystyle \lim_{x \to 0}\frac{x}{\sin x} = 1 $ you have to show that $\displaystyle \lim_{x \to 0}(\frac{\sin x}{x})'=0$. Use l'Hospital rule again:

$$\lim_{x \to 0}(\frac{\sin x}{x})'=\lim_{x \to 0}\frac{x\cos x-\sin x}{x^2}=^{H}\lim_{x \to 0} \frac{-x\sin x +\cos x-\cos x}{2x}=\lim_{x \to 0}-\frac{1}{2}\sin x=0$$

Help Evaluating $\lim_{x\to0}(\frac{\sin{x}}{{x}})^{\frac{1}{x}}$

4 Answers4