Question

Find the value of limx → 0⁡(Sin(2x))^Tan^2 (2x)?

(a) e^0.5

(b) e^-0.5

(c) e^-1

(d) e

This question was addressed to me in a national level competition.

I would like to ask this question from Indeterminate Forms in section Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (b) e^-0.5

Easiest explanation: y=\(\lim_{x\rightarrow 0}(sin(2x))^{tan^2(2x)}\)

Taking log of both side

\(ln y=\lim_{x\rightarrow 0}\frac{ln(sin(2x))}{cot^2(2x)}(0/0)\)

By L’Hospital Rule

\(ln y=-\lim_{x\rightarrow 0}\frac{2cos(2x)}{sin(2x).4.cosec^2(2x)cot(2x)}=-0.5\lim_{x\rightarrow 0}sin^2(2x)\)=-0.5

=>y=e^-0.5