Question

If \[ f(x)=\frac{e^{1/x}-1}{e^{1/x}+1}, \] then

• A. \[ \lim_{x\to 0}f(x)=0 \]
• B. \[ \lim_{x\to \infty}f(x)=1 \]
• C. \[ \lim_{x\to 0}f(x)=-1 \]
• D. \[ \lim_{x\to \infty}f(x)=0 \]

Hint:

For expressions involving \(e^{1/x}\), always check the limits from the left and right separately when \(x\to 0\), because \(1/x\) approaches \(+\infty\) and \(-\infty\) from opposite sides.

Answer 1

The Correct Option is D

Step 1: Evaluate \(\displaystyle \lim_{x\to\infty}f(x)\).
Given \[ f(x)=\frac{e^{1/x}-1}{e^{1/x}+1}. \] As \[ x\to\infty, \] we have \[ \frac1x\to 0. \] Therefore, \[ e^{1/x}\to e^0=1. \] Substituting, \[ \lim_{x\to\infty}f(x) = \frac{1-1}{1+1} = 0. \] Hence, \[ \boxed{\lim_{x\to\infty}f(x)=0}. \] So option (4) is true.

Step 2: Check option (2).
From Step 1, \[ \lim_{x\to\infty}f(x)=0, \] not \(1\). Therefore option (2) is false.

Step 3: Check the limit as \(x\to 0\).
When \[ x\to 0^+, \] we have \[ \frac1x\to +\infty. \] Thus, \[ e^{1/x}\to \infty, \] and \[ f(x) = \frac{e^{1/x}-1}{e^{1/x}+1} \to 1. \] When \[ x\to 0^-, \] we have \[ \frac1x\to -\infty. \] Thus, \[ e^{1/x}\to 0, \] and \[ f(x) = \frac{0-1}{0+1} =-1. \] Hence, \[ \lim_{x\to 0^+}f(x)=1, \qquad \lim_{x\to 0^-}f(x)=-1. \] Since the left-hand and right-hand limits are different, \[ \lim_{x\to 0}f(x) \] does not exist.

Step 4: Verify the remaining options.
Since \[ \lim_{x\to 0}f(x) \] does not exist, neither \[ \lim_{x\to 0}f(x)=0 \] nor \[ \lim_{x\to 0}f(x)=-1 \] is true.

Step 5: Final conclusion.
The only correct statement is \[ \boxed{\lim_{x\to\infty}f(x)=0}. \] Therefore, the correct answer is \[ \boxed{(4)}. \]