To determine the value of \(k\) such that the function \(f(x)\) given by:
\(f(x) = \begin{cases} (\cos x)^{1/x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases}\)
is continuous at \(x = 0\), we first need to evaluate the limit of \(f(x)\) as \(x\) approaches 0:
\(\lim_{{x \to 0}} (\cos x)^{1/x^2}\)
For continuity at \(x = 0\), this limit must be equal to \(k\):
\(\lim_{{x \to 0}} (\cos x)^{1/x^2} = k\)
Firstly, consider the expression \((\cos x)^{1/x^2}\). Taking the natural logarithm, we have:
\(\ln((\cos x)^{1/x^2}) = \frac{1}{x^2} \ln(\cos x)\)
To find the limit, we'll evaluate:
\(\lim_{{x \to 0}} \frac{\ln(\cos x)}{x^2}\)
As \(x \to 0\), \(\cos x \to 1\), and thus \(\ln(\cos x) \approx -\frac{x^2}{2}\) using the expansion \(\ln(1-y) \approx -y\):
\(\ln(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}\)
Substituting back, we have:
\(\lim_{{x \to 0}} \frac{-\frac{x^2}{2}}{x^2} = \lim_{{x \to 0}} -\frac{1}{2} = -\frac{1}{2}\)
Exponentiating to solve for the original expression:
\(\lim_{{x \to 0}} (\cos x)^{1/x^2} = e^{-\frac{1}{2}}\)
Thus, the function is continuous at \(x = 0\) if:
\(k = e^{-\frac{1}{2}}\)
Given the options: 8, 1, -1, None of these, the value of \(k\) does not match any of these, confirming the correct answer is None of these.