Question

Let A = \( \lim_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}} \), then log\(_e\) A =}

• A. 2
• B. 1
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{4} \)

Hint:

$\lim (1 + \lambda)^{\frac{1}{\lambda}} = e$.

Answer 1

The Correct Option is C

Step 1: Form $1^\infty$
$\lim_{x \to 0^+} f(x)^{g(x)} = e^{\lim [f(x)-1]g(x)}$.
Limit $= e^{\lim_{x \to 0} (\tan^2 \sqrt{x}) \cdot \frac{1}{2x}}$.
Step 2: Simplify exponent
$\lim_{x \to 0} \frac{\tan^2 \sqrt{x}}{2x} = \frac{1}{2} \lim_{x \to 0} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)^2$.
Since $\lim_{u \to 0} \frac{\tan u}{u} = 1$, the exponent is $\frac{1}{2}(1)^2 = \frac{1}{2}$.
Step 3: Conclusion
$A = e^{1/2} \implies \log_e A = \frac{1}{2}$.
Final Answer:(C)