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: Understanding the Concept:
Near \(x=0\) we use standard limits: \(\dfrac{a^x-1}{x}\to\ln a\), \(\dfrac{\tan x}{x}\to1\) and \(\dfrac{\ln(1+x)}{x}\to1\).

Step 2: Rewrite:
\[ \frac{(4^x-1)^3}{\tan\frac x4\,\ln\left(1+\frac{x^2}{3}\right)}=\frac{\left(\frac{4^x-1}{x}\right)^3 x^3}{\frac{\tan(x/4)}{x/4}\cdot\frac x4\cdot\frac{\ln(1+x^2/3)}{x^2/3}\cdot\frac{x^2}3} \]

Step 3: Take the limit:
The numerator tends to \((\ln4)^3x^3\). The denominator tends to \(1\cdot\frac x4\cdot1\cdot\frac{x^2}{3}=\frac{x^3}{12}\). So
\[ L=12(\ln4)^3=12(2\ln2)^3=96(\ln2)^3 \]

Step 4: Match:
Compare with \(96(\log a)^b\): \(a=2\), \(b=3\). So \(a+b=5\), option (A).

Final Answer:
The limit is 96 (ln 2)^3, so a = 2, b = 3 and a + b = 5. \[ \boxed{5} \]