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: Understanding the Concept:
We need the standard limits \(\lim_{x\to0}\frac{\log(1+ax)}{x} = a\) and \(\lim_{x\to0}\frac{e^{ax}-1}{x} = a\).

Step 2: Rewrite the expression:
\[ \frac{x\log(1+4x)}{(e^{4x}-1)^2} = \frac{\log(1+4x)}{x} \cdot \frac{x^2}{(e^{4x}-1)^2} = \frac{\log(1+4x)}{x} \cdot \left(\frac{x}{e^{4x}-1}\right)^2 \]

Step 3: Apply limits:
\[ \lim = 4 \times \left(\frac{1}{4}\right)^2 = 4 \times \frac{1}{16} = \frac14 \]

Step 4: Why the other options are wrong.
\(\frac1{16}\) comes from forgetting the factor 4 from the logarithm. \(\frac13\) and \(\frac19\) do not follow from these standard limits.

Final Answer:
The limit is \(\frac14\), option (A). \[ \boxed{\frac{1}{4}} \]