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 the limit as x to +infinity.
As \(x\to+\infty\), \(\frac{1}{x}\to0^+\), so \(e^{1/x}\to1^+\). Then \(f(x)=\frac{e^{1/x}-1}{e^{1/x}+1}\to\frac{0}{2}=0\).

Step 2: Check other options briefly.
As \(x\to0^+\): \(e^{1/x}\to\infty\), \(f\to1\) (not \(0\)). As \(x\to0^-\): \(e^{1/x}\to0\), \(f\to\frac{-1}{1}=-1\). So \(\lim_{x\to0}f(x)\) does not exist. \[\boxed{\lim_{x\to\infty}f(x)=0}\]