How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$?

Solution 1:

For small values of $x$, the Taylor series for $\sin(x)$ is $x - x^3 / 6$; then, for $\sin(x)/ x$, the expansion around $x=0$ is $1 - x^2 /6$. Now, remember that, for small values of $y$, $(1+y)^a$ is approximated by $1 + a y$ (another Taylor series). So, for your expression, you arrive to

$$(1 - x^2 /6)^{1/x}$$

which is almost $(1 - x / 6)$. So, your limit is $1$.

Edit

Probably better (at least to me three years after my first answer), considering $$ A=\left(\frac{ \sin x}{x}\right)^{1/x}\implies \log(A)=\frac 1x \log\left(\frac{ \sin x}{x}\right)$$ Now, using Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\frac{ \sin x}{x}=1-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log\left(\frac{ \sin x}{x}\right)=-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(A)=\frac 1x \log\left(\frac{ \sin x}{x}\right)=-\frac{x}{6}+O\left(x^3\right)$$ Now, using $A=e^{\log(A)}$ and Taylor again $$A=1-\frac{x}{6}+\frac{x^2}{72}+O\left(x^3\right)$$ Unsing $x=\frac \pi 6$, the "exact" value $$\left(\frac{3}{\pi }\right)^{6/\pi }\approx 0.915689$$ while the above asymptotics would give $$1-\frac{\pi }{36}+\frac{\pi ^2}{2592}\approx 0.916541$$

Solution 2:

Hint : Take its logarithm, use the fact that $\ln a^b=b\ln a=\dfrac{\ln a}{1/b}$, and apply l'Hopital $3$ or $4$ times.

$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$

$$=\lim_{x\to0}\frac{(1\cdot\cos x-x\cdot\sin x)-\cos x}{1\cdot\sin x+x\cdot\cos x}=-\lim_{x\to0}\frac{x\cdot\sin x}{\sin x+x\cdot\cos x}=$$

$$=-\lim_{x\to0}\frac{1\cdot\sin x+x\cdot\cos x}{\cos x+(1\cdot\cos x-x\cdot\sin x)}=-\lim_{x\to0}\frac{\sin x+x\cdot\cos x}{2\cdot\cos x-x\cdot\sin x}=-\frac{0+0\cdot0}{2\cdot1-0\cdot0}=-\frac02$$

$$=0\iff L=e^0=1.$$

Solution 3:

Let $$ y = \frac{\sin x}{x}-1 $$ and notice that for $x\to 0$ also $y\to 0$. Then remember that $$ \lim_{y\to 0} \left(1+y\right)^{\frac 1 y} = e $$ while (use Hopital or Taylor here) $$ \frac{y}{x} = \frac{\frac{\sin x}{x} - 1}{x} = \frac{\sin x - x}{x^2} \to 0. $$ So your limit is $$ (1+y)^\frac 1x = \left((1+y)^\frac{1}{y}\right)^{\frac{y}{x}} \to e^0=1 $$

Solution 4:

Notice by Young's Inequality since $(\frac{1}{x} + (1 - \frac{1}{x})) = 1$, then

$$ (\frac{\sin x}{x} )^{1/x}=(\frac{\sin x}{x} )^{1/x} 1^{1 - \frac{1}{x}} \leq \frac{\sin x}{x^2} + 1 - \frac{1}{x} = \frac{\sin x - x}{x^2} + 1$$

Now, for positive $x$ and for $a \leq 1$, we have that $a^x \leq x +1 $. Hence $a^{1/x} \geq \frac{1}{x+1}$. Now, since $\frac{\sin x}{x} \leq 1$, we apply this inequality with $a = \frac{\sin x}{x} $ to obtain

$$ (\frac{\sin x}{x} )^{1/x} \geq \frac{1}{x+1}$$. Hence we have

$$ \frac{1}{x+1} \leq (\frac{\sin x}{x} )^{1/x} \leq \frac{\sin x - x}{x^2} + 1$$.

Now, since $\lim{ \frac{1}{x+1} } = 1 $ and

$$ \lim (\frac{\sin x - x}{x^2} + 1 ) = \lim ( \frac{\sin x - x}{x^2} ) = 1 + \lim ( \frac{\cos x - 1}{2x} ) = 1 + \lim ( \frac{- \sin x}{2} ) = 1 $$

Now, result follows by the squeeze trick.