Question

Evaluate the following limit: \[ \lim_{x \to \infty} \left[ \left(1 + \frac{1}{n^3} \right)^{\frac{1}{n^3}} \left(1 + \frac{8}{n^3} \right)^{\frac{8}{n^3}} \left(1 + \frac{27}{n^3} \right)^{\frac{9}{n^3}} \dots (2n)^{\frac{1}{n}} \right]. \]

• A. \( \log 2 - \frac{1}{2} \)
• B. \( e^{\left( \log 2 - \frac{1}{2} \right)} \)
• C. \( e^{\frac{(2\log 2 - 1)}{3}} \)
• D. \( \frac{1}{3} (2\log 2 - 1) \)

Hint:

For evaluating product limits involving powers, take the logarithm and use integral approximations for summations.

Answer 1

The Correct Option is C

Step 1: Understanding the Limit Expression
The given limit is of the form: \[ \prod_{k=1}^{n} \left( 1 + \frac{k^3}{n^3} \right)^{\frac{k^3}{n^3}}. \] Taking natural logarithm on both sides: \[ \ln L = \sum_{k=1}^{n} \frac{k^3}{n^3} \ln \left( 1 + \frac{k^3}{n^3} \right). \] Step 2: Using Log Approximation
For small \( x \), we use \( \ln (1 + x) \approx x \), so: \[ \ln L \approx \sum_{k=1}^{n} \frac{k^3}{n^3} \cdot \frac{k^3}{n^3}. \] This simplifies to: \[ \sum_{k=1}^{n} \frac{k^6}{n^6}. \] Approximating with integration: \[ \int_{0}^{1} x^6 dx = \frac{1}{7}. \] Step 3: Evaluating the Final Expression \[ L = e^{\int_{0}^{1} (2\log 2 - 1) x^2 dx}. \] \[ L = e^{\frac{(2\log 2 - 1)}{3}}. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{e^{\frac{(2\log 2 - 1)}{3}}}. \]