Question

If \(\underset{x\rightarrow 0}{lim}\frac{(4^x-1)^3}{tan(\frac{x}{4})log(1+\frac{x^2}{3})} = 96(loga)^b\), then \((a+b) =\)

• A. \(5\)
• B. \(7\)
• C. \(3\)
• D. \(4\)

Hint:

Use small x approximations for each factor.

Answer 1

The Correct Option is A

Step 1: Series approach:
For small $x$: $4^x-1\approx x\ln4$, $\tan\frac x4\approx\frac x4$, $\ln\left(1+\frac{x^2}3\right)\approx\frac{x^2}3$.

Step 2: Ratio:
\[ \frac{x^3(\ln4)^3}{\frac x4\cdot\frac{x^2}{3}}=12(\ln4)^3 \]

Step 3: Simplify:
$\ln4=2\ln2$, so $12\cdot8(\ln2)^3=96(\ln2)^3$.

Step 4: Read a and b:
$a=2$, $b=3$, giving $5$.

Final Answer:
Compare 12 (ln 4)^3 with 96 (log a)^b. \[ \boxed{A} \]