Question

A function \( f(x) \) is given by:
\[ f(x) = \begin{cases} \frac{1}{e^x - 1}, & \text{if } x \neq 0 \\ \frac{1}{e^x + 1}, & \text{if } x = 0 \end{cases} \] Then, which of the following is true?

• A. not continuous at x = 0
• B. differentiable at x = 0
• C. differentiable at x = 0, but not continuous at x = 0
• D. continuous at x = 0

Hint:

When dealing with piecewise functions, always check the limit from both sides and compare it with the function's value at the point of interest. If they don't match, the function is not continuous there.

Answer 1

The Correct Option is A

The function is given as:

\[ f(x) = \begin{cases} \frac{1}{e^x - 1}, & \text{if } x \neq 0 \\ \frac{1}{e^x + 1}, & \text{if } x = 0 \end{cases} \]

We need to check whether the function is continuous at \( x = 0 \) and differentiable at \( x = 0 \).
1. Continuity check at \( x = 0 \):
The function is continuous at a point if the limit of the function as \( x \) approaches that point is equal to the function's value at that point. - For \( x \neq 0 \), we have: \[ \lim_{x \to 0} \frac{1}{e^x - 1}. \] Since \( e^x \to 1 \) as \( x \to 0 \), the denominator approaches \( 0 \), and thus the limit does not exist. Thus, the function is not continuous at \( x = 0 \), and the correct answer is option (1).