Question

\(\underset{x\rightarrow 0}{lim}[\frac{x\cdot log(1+4x)}{(e^{4x}-1)^2}] = \cdots\)

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{16}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{1}{9}\)

Hint:

Use log(1+4x) ~ 4x and (e^{4x}-1) ~ 4x as x tends to 0.

Answer 1

The Correct Option is A

Step 1: Replace by equivalents near 0:
$\log(1+4x) \approx 4x$ and $e^{4x} - 1 \approx 4x$.

Step 2: Substitute:
\[ \frac{x \cdot 4x}{(4x)^2} = \frac{4x^2}{16x^2} = \frac14 \]

Step 3: Check:
Both numerator and denominator are of order $x^2$, so the limit is a finite constant, 1/4.

Final Answer:
The limit is 1/4, option (A). \[ \boxed{\frac{1}{4}} \]