Question

If the function \[ f(x) = \begin{cases} (\cos x)^{1/x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \] is continuous at \( x = 0 \), then the value of \( k \) is:

• A. 8
• B. 1
• C. \(-1\)
• D. None of these

Hint:

Use exponential form: \(a^b = e^{b \ln a}\) for limits of the form \(1^\infty\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For continuity, \(\lim_{x \to 0} f(x) = f(0) = k\).

Step 2: Detailed Explanation:
\(\lim_{x \to 0} (\cos x)^{1/x^2} = e^{\lim_{x \to 0} \frac{\ln(\cos x)}{x^2}}\)
\(\ln(\cos x) = \ln\left(1 - \frac{x^2}{2} + \cdots\right) = -\frac{x^2}{2} - \frac{x^4}{12} + \cdots\)
\(\frac{\ln(\cos x)}{x^2} \to -\frac{1}{2}\)
So limit = \(e^{-1/2} = \frac{1}{\sqrt{e}}\).
Thus \(k = \frac{1}{\sqrt{e}}\), which is not among the options.

Step 3: Final Answer:
Option (D) None of these.