2 votes 2 votes $\lim_{x\rightarrow a} f(x)^{g(x)} = e^{\lim_{x\rightarrow a}g(x)[f(x)-1]}$ Solve the below limit without using the above formula, $\lim_{x \rightarrow 0} ({\frac{sin x}{x}})^{\frac{sin x}{x - sin x}}$ dutta007sourish asked Sep 7, 2021 dutta007sourish 337 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply ankitgupta.1729 commented Sep 9, 2021 reply Follow Share Answer should be $\frac{1}{e}$ $y=\lim_{x\rightarrow 0} (\frac{\sin x}{x})^{\frac{\sin x}{x – \sin x}}$ Take log both sides, $\ln y = \lim_{x\rightarrow 0} \left(\frac{\sin x}{x – \sin x} \right) \ln (\frac{\sin x}{x})$ $\ln y = \lim_{x\rightarrow 0} \frac{ \left (\sin x \ln (\frac{\sin x}{x}) \right)}{x – \sin x}$ Since, $\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1$, So, you get 0/0 form and so, apply L’Hopital’s rule in right side, you will get, $\ln y= \lim_{x \rightarrow 0} \frac{\left(\cos x \ln (\frac{\sin x}{x}) + \cos x – \frac{\sin x }{x} \right)}{1- \cos x}$ Here, you get 0/0 form again in right side, so apply L’Hopital’s rule again, you will get, $\ln y = \lim_{x \rightarrow 0} \left( – \ln (\frac{\sin x}{x}) + \frac{x\cos^2 x – \sin x \cos x}{x \sin^2 x} – 1 – \frac{x\cos x – \sin x}{x^2 \sin x}\right)$ Since, $\lim_{x \rightarrow 0} \frac{x\cos^2 x – \sin x \cos x}{x \sin^2 x} = \frac{-1}{3} $ by L’Hopital’s rule and $\lim_{x \rightarrow 0} \frac{x\cos x – \sin x}{x^2 \sin x} = \frac{-1}{3}$ by L’Hopital’s rule. So, you’ll get $\ln y = -1$ which means $y = \frac{1}{e}$ 2 votes 2 votes dutta007sourish commented Sep 9, 2021 reply Follow Share Thanks, Ankit 0 votes 0 votes shuham kumar commented Sep 11, 2022 reply Follow Share my ansr is comming 0 0 votes 0 votes [ Jiren ] commented Sep 12, 2022 reply Follow Share @ankitgupta.1729 sir Can we say this is of the form 1^infinity as x tends to 0 0 votes 0 votes ankitgupta.1729 commented Sep 12, 2022 reply Follow Share yes. 0 votes 0 votes Please log in or register to add a comment.