Question

\(\underset{x\rightarrow 0}{lim}\frac{(5^x-1)^4\,\text{cosec}\,(xlog5)}{tan(xlog5)\cdot log(1+x^2log25)} = \ldots \ldots\)

• A. \(5log5\)
• B. \(log\sqrt{5}\)
• C. \((log5)^2\)
• D. \(\frac{1}{4}log5\)

Hint:

Replace each factor by its small-x approximation.

Answer 1

The Correct Option is B

Step 1: Standard limits:
Use $\lim \dfrac{a^x-1}{x} = \log a$, $\lim\dfrac{\tan u}{u} = 1$, $\lim\dfrac{\sin u}{u}=1$ and $\lim\dfrac{\log(1+u)}{u} = 1$.

Step 2: Rearrange:
Write $\operatorname{cosec}(x\log5) = 1/\sin(x\log5)$. The limit is $\dfrac{\left(\frac{5^x-1}{x}\right)^4 x^4}{\sin(x\log5)\tan(x\log5)\cdot \log(1+x^2\log25)}$.

Step 3: Collect powers of x:
Numerator $\to (\log5)^4 x^4$. Denominator $\to (x\log5)(x\log5)(x^2\log25) = x^4(\log5)^2\cdot2\log5$. Ratio $= \dfrac{(\log5)^4}{2(\log5)^3} = \dfrac{\log5}{2}$, option (B).

Final Answer:
The limit is log of root 5. \[ \boxed{\text{(B) }\log\sqrt5} \]