Question

If \[ \lim_{x\to0}\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=k, \] then \(e^{2k}=\)

• A. \(1\)
• B. \(\log 2\)
• C. \(2\)
• D. \(\dfrac12\log2\)

Hint:

For limits involving \[ a^{u(x)}-a^{v(x)}, \] use \[ \boxed{ a^{u}-a^{v} \approx a^v\log(a)\,(u-v) } \] for small values of \[ u-v. \]

Answer 1

The Correct Option is C

Step 1: Apply the Mean Value Theorem. Let \[ f(t)=2^t. \] Then \[ 2^{\tan x}-2^{\sin x} = 2^\xi\log2\,(\tan x-\sin x), \] where \[ \xi \] lies between \[ \sin x \] and \[ \tan x. \] As \[ x\to0, \] \[ 2^\xi\to1. \] Hence, \[ k = \log2\, \lim_{x\to0} \frac{\tan x-\sin x}{x^2\sin x}. \]

Step 2: Evaluate the limit. Using \[ \tan x-\sin x = \frac{\sin x(1-\cos x)}{\cos x}, \] we get \[ \frac{\tan x-\sin x}{x^2\sin x} = \frac{1-\cos x}{x^2\cos x}. \] Now, \[ \lim_{x\to0} \frac{1-\cos x}{x^2} = \frac12, \qquad \lim_{x\to0}\cos x=1. \] Therefore, \[ k = \frac{\log2}{2}. \]

Step 3: Find \(e^{2k}\). Hence, \[ e^{2k} = e^{\log2} = 2. \] Therefore, \[ \boxed{2}. \] Thus, \[ \boxed{(C)} \] is the correct answer.