Mathematics - Limits Question with Solution | TestHub

The limit is of the form $1^{\infty }$. We can evaluate it as $e^{{\lim }_{x\to 0}\left(\frac{\sin x}{x}-1\right)\frac{1}{x^2}}$. The exponent is ${\lim }_{x\to 0}\frac{\sin x-x}{x^3}$. Applying L'Hopital's rule three times: ${\lim }_{x\to 0}\frac{\cos x-1}{3x^2}={\lim }_{x\to 0}\frac{-\sin x}{6x}={\lim }_{x\to 0}\frac{-\cos x}{6}=-\frac{1}{6}$. Therefore, the value of the limit is $e^{-1/6}$.