Question

Evaluate \[ \lim_{x\to0}\frac{e^{\tan x}-e^{\sin x}}{(1-\cos x)\tan x} \]

• A. \(0\)
• B. \(\dfrac12\)
• C. \(1\)
• D. Does not exist

Hint:

For limits involving exponential functions, \[ \boxed{ e^a-e^b\sim a-b \quad(a,b\to0) } \] This approximation often simplifies complicated limits.

Answer 1

The Correct Option is C

Step 1: Apply the Mean Value Theorem to the exponential function. For the function \[ f(t)=e^t, \] there exists a number \(\xi\) between \(\sin x\) and \(\tan x\) such that \[ e^{\tan x}-e^{\sin x} = e^\xi(\tan x-\sin x). \] As \[ x\to0, \] \[ \xi\to0 \] and therefore, \[ e^\xi\to1. \] Hence, \[ \lim_{x\to0} \frac{e^{\tan x}-e^{\sin x}} {(1-\cos x)\tan x} = \lim_{x\to0} \frac{\tan x-\sin x} {(1-\cos x)\tan x}. \]

Step 2: Simplify the numerator. Using \[ \tan x=\frac{\sin x}{\cos x}, \] we get \[ \tan x-\sin x = \sin x\left(\frac1{\cos x}-1\right) = \frac{\sin x(1-\cos x)}{\cos x}. \] Hence, \[ \frac{\tan x-\sin x} {(1-\cos x)\tan x} = \frac{\sin x}{\cos x\tan x}. \] Since \[ \tan x=\frac{\sin x}{\cos x}, \] \[ \frac{\sin x}{\cos x\tan x}=1. \] Therefore, \[ \lim_{x\to0} \frac{e^{\tan x}-e^{\sin x}} {(1-\cos x)\tan x} =1. \] Thus, \[ \boxed{1} \] is the correct answer. Hence, \[ \boxed{(C)} \] is the correct answer.