Question

\(\mathop {\lim }\limits_{x \to 0} \frac{\left(7^x-1\right)^4}{\tan \left(\frac{x}{\mathrm{k}}\right) \cdot \log \left(1+\frac{x^2}{3}\right) \cdot \sin 4 x}=3(\log 7)^3\), then \(\mathrm{k}=\)

• A. \(4(\mathrm{log}7)^{-1}\)
• B. \(\frac{1}{4}(\mathrm{log}7)^{-1}\)
• C. \(4\mathrm{log}(\frac{1}{7})\)
• D. \(\frac{1}{4}\mathrm{log}7\)

Hint:

Use the standard limits for (a^x - 1)/x, tan u/u, log(1+u)/u and sin u/u.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Divide and multiply by powers of \(x\) so that each factor takes a standard form as \(x\to0\).

Step 2: Numerator
\(\left(7^x-1\right)^4\sim(x\log7)^4=x^4(\log7)^4\).

Step 3: Denominator
\(\tan\frac xk\sim\frac xk\), \(\log\left(1+\frac{x^2}{3}\right)\sim\frac{x^2}{3}\), \(\sin4x\sim4x\), so the product is \(\frac{4x^4}{3k}\).

Step 4: Equate
Limit \(=\frac{3k(\log7)^4}{4}=3(\log7)^3\Rightarrow k=\frac{4}{\log7}=4(\log7)^{-1}\).

Final Answer:
\(k=4(\log7)^{-1}\). \[ \boxed{\text{(A) }4(\log 7)^{-1}} \]