Question

If \[ f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}, \quad x \neq 0 \quad \text{and} \quad f(0) = 0, \] then \(f(x)\) is:

• A. left continuous at 0
• B. right continuous at 0
• C. discontinuous at 0
• D. continuous at 0

Hint:

For functions with \(e^{1/x}\), check left and right limits separately as \(x \to 0^+\) and \(x \to 0^-\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Check left-hand and right-hand limits at \(x=0\).

Step 2: Detailed Explanation:
As \(x \to 0^+\), \(1/x \to +\infty\), \(e^{1/x} \to \infty\), so \(f(x) \to \frac{\infty - 1}{\infty + 1} = 1\).
As \(x \to 0^-\), \(1/x \to -\infty\), \(e^{1/x} \to 0\), so \(f(x) \to \frac{0 - 1}{0 + 1} = -1\).
LHL = \(-1\), RHL = \(1\), and \(f(0)=0\).
Thus, not continuous at 0.

Step 3: Final Answer:
Option (C) discontinuous at 0.