Question

$\lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x} =$}

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Hint:

If you see $e^A - e^B$, always factor out $e^B$ to create the form $e^{A-B} - 1$.

Answer 1

The Correct Option is A

Step 1: Concept
This is a limit of the form $\frac{0}{0}$. Use the standard limit identity $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$.

Step 2: Meaning
Factor out $e^x$ from the numerator.

Step 3: Analysis
$\lim_{x \to 0} \frac{e^x(e^{\tan x - x} - 1)}{\tan x - x}$. As $x \to 0$, $(\tan x - x) \to 0$. The limit becomes: $(\lim_{x \to 0} e^x) \times (\lim_{x \to 0} \frac{e^{\tan x - x} - 1}{\tan x - x})$. $= e^0 \times 1 = 1 \times 1 = 1$.

Step 4: Conclusion
The limit evaluates to 1. Final Answer: (A)