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

To determine the nature of the function \(f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}\) at \(x = 0\), we need to assess its continuity at that point. According to the problem, \(f(0) = 0\).

1. For a function to be continuous at a point \(x = 0\), it must satisfy the condition: \(\lim_{{x \to 0}} f(x) = f(0)\).
2. Let’s compute \(\lim_{{x \to 0}} f(x)\):

Substitute \(y = \frac{1}{x}\), as \(x \to 0\) implies \(y \to \infty\).

1. Now we examine \(\lim_{{y \to \infty}} \frac{e^{y} - 1}{e^{y} + 1}\):
2. Divide numerator and denominator by \(e^{y}\) to simplify: \(\frac{e^{y} - 1}{e^{y} + 1} = \frac{1 - e^{-y}}{1 + e^{-y}}\).
3. As \(y \to \infty\), \(e^{-y} \to 0\). Hence, the expression simplifies to: \(\frac{1 - 0}{1 + 0} = 1\).

The limit of \(f(x)\) as \(x \to 0\) is 1, but we have \(f(0) = 0\). Therefore:

• \(\lim_{{x \to 0}} f(x) \neq f(0)\),
• This means \(f(x)\) is not continuous at \(x = 0\).

Conclusion: The function \(f(x)\) is discontinuous at \(x = 0\), making the correct answer "discontinuous at 0".