Question

Let the function \(f(x)\) be defined as:
\(f(x) = \left\{ \begin{array}{cc}[tan(\frac{π}{4}+x)]^{\frac{1}{x}}, & x\neq 0 \\ k, & x = 0\end{array} \right.\)
If \(f(x)\) is continuous at \(x = 0\), then the value of \(k\) is...

• A. \(e\)
• B. \(e^2\)
• C. \(\frac{1}{e^2}\)
• D. \(\frac{1}{e}\)

Hint:

Continuity needs k to equal the limit. The limit is of 1 to the power infinity form, so take the log and differentiate.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
For \(f\) to be continuous at \(x = 0\), we need \(k = \lim_{x\to0} f(x)\).

Step 2: Identify the form:
As \(x \to 0\), \(\tan(\pi/4 + x) \to \tan(\pi/4) = 1\) and the power \(1/x \to \infty\). This is the \(1^{\infty}\) form, so we use logarithms.

Step 3: Take the logarithm:
Let \(L = \lim [\tan(\pi/4 + x)]^{1/x}\). Then \[ \ln L = \lim_{x\to0} \frac{\ln\tan(\pi/4 + x)}{x} \] This is of the form 0/0, since \(\ln\tan(\pi/4) = \ln 1 = 0\).

Step 4: Differentiate (L'Hopital's rule):
The derivative of \(\ln\tan(\pi/4 + x)\) is \(\frac{\sec^2(\pi/4 + x)}{\tan(\pi/4 + x)}\). At \(x = 0\): \[ \frac{\sec^2(\pi/4)}{\tan(\pi/4)} = \frac{2}{1} = 2 \]

Step 5: Finish:
So \(\ln L = 2\), and \(L = e^2\). Therefore \(k = e^2\).

Step 6: Cross-check:
Near \(x=0\), \(\tan(\pi/4 + x) \approx 1 + 2x\). Then \((1 + 2x)^{1/x} \to e^2\). The options \(e\), \(1/e\) and \(1/e^2\) would need slopes 1, -1 and -2, which do not match.

Final Answer:
The value of \(k\) is \(e^2\), option (B). \[ \boxed{k=e^2} \]