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

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.

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:

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

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

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

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:

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

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

• It might be more helpful to the OP if you specified where you applied L'Hopital's rule! Nonetheless, nice solution! – Kari 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! – Kari 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