## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{1/x^2}=e^{-\frac{1}{6}}.$$
To solve this limit we will use the fact that the natural exponential function and the natural logarithmic function are inverse to each other i.e. that $e^{\ln a}=a$. This gives $\left(\frac{\sin x}{x}\right)^{1/x^2}=e^{\ln\left(\frac{\sin x}{x}\right)^{1/x^2}}=e^{\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)}.$ Since the natural exponential function is continuous it can exchange places with the limit so we get: $$L=\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{1/x^2}=\lim_{x\to0}e^{\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)}=e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)}=e^l,$$ where we have denoted $l=\lim_{x\to0}\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)$. Now we have to solve this limit. LR will stand for Apply L'Hopital's rule: $$l=\lim_{x\to0}\frac{1}{x^2}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to0}\frac{\ln\left(\frac{\sin x}{x}\right)}{x^2}=\left[\frac{\ln\left(\frac{\sin 0}{0}\right)}{0^2}\right]=\left[\frac{\ln 1}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{\left(\ln\left(\frac{\sin x}{x}\right)\right)'}{(x^2)'}=\lim_{x\to0}\frac{\frac{1}{\frac{\sin x}{x}}\left(\frac{\sin x}{x}\right)'}{2x}=\lim_{x\to0}\frac{\frac{x}{\sin x}\frac{(\sin x)'x-\sin x (x)'}{x^2}}{2x}=\lim_{x\to0}\frac{\frac{x}{\sin x}\frac{x\cos x-\sin x}{x^2}}{2x}=\lim_{x\to0}\frac{\frac{x\cos x-\sin x}{x\sin x}}{2x}=\lim_{x\to0}\frac{x\cos x-\sin x}{2x^2\sin x}=\left[\frac{0\cdot\cos0-\sin 0}{2\cdot0^2\cdot\sin 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(x\cos x-\sin x)'}{(2x^2\sin x)'}=\lim_{x\to0}\frac{(x)'\cos x+ x(\cos x)'-(\sin x)'}{2((x^2)'\sin x+x^2(\sin x)')}=\lim_{x\to0}\frac{\cos x-x\sin x-\cos x}{2(2x\sin x+x^2\cos x)}=\lim_{x\to0}\frac{-x\sin x}{4x\sin x+2x^2\cos x}=\lim_{x\to0}\frac{-\sin x}{4\sin x + 2x\cos x}=\left[\frac{-\sin 0}{4\sin 0+2\cdot0\cdot\cos 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(-\sin x)'}{(4\sin x+x\cos x)'}=\lim_{x\to0}\frac{-\cos x}{4\cos x+2(x)'\cos x+2x(\cos x)'}=\lim_{x\to0}\frac{-\cos x}{4\cos x+2\cos x-2x\sin x}=\lim_{x\to0}\frac{-\cos x}{6\cos x-2x\sin x}=\frac{-\cos 0}{6\cos 0-2\cdot0\cdot\sin 0}=-\frac{1}{6}.$$ Putting this into the initial lmit: $$L=e^l=e^{-\frac{1}{6}}.$$