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: Understanding the Concept:
Near \(x = 0\): \(5^x - 1 \approx x\log 5\), \(\tan(x\log5) \approx x\log5\), \(\operatorname{cosec}(x\log5) \approx \dfrac{1}{x\log5}\), and \(\log(1 + u) \approx u\).

Step 2: Replace each part:
Numerator: \((x\log5)^4 \cdot \dfrac{1}{x\log5} = (x\log5)^3\).
Denominator: \(x\log5 \cdot x^2\log25 = x\log5\cdot x^2\cdot 2\log5 = 2x^3(\log5)^2\).

Step 3: Divide:
\[ \frac{x^3(\log5)^3}{2x^3(\log5)^2} = \frac{\log5}{2} \]

Step 4: Match the option:
\(\dfrac{\log5}{2} = \log 5^{1/2} = \log\sqrt5\), which is option (B).

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