Question:
\[
\lim_{x\to0}\frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}+1}=\,\_
\]
\[
\lim_{x\to0}\frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}+1}=\,\_
\]
Show Hint
For $e^{1/x}$, always check both sides of zero! It changes behavior from infinite to zero.
Updated On: Jul 4, 2026
- $e^{0}$
- 1
- 0
- does not exist
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Concept A limit exists only if the left-hand limit (LHL) and right-hand limit (RHL) are equal.
Step 2: Meaning As $x \to 0^+$, $1/x \to \infty$, so $\frac{e^{\infty}}{e^{\infty}+1}$ behaves like $\frac{e^{\infty}}{e^{\infty}} = 1$.
Step 3: Analysis As $x \to 0^-$, $1/x \to -\infty$, so $e^{1/x} \to 0$. The limit becomes $\frac{0}{0+1} = 0$.
Step 4: Conclusion Since RHL (1) $\ne$ LHL (0), the limit does not exist. Final Answer: (D)
Step 2: Meaning As $x \to 0^+$, $1/x \to \infty$, so $\frac{e^{\infty}}{e^{\infty}+1}$ behaves like $\frac{e^{\infty}}{e^{\infty}} = 1$.
Step 3: Analysis As $x \to 0^-$, $1/x \to -\infty$, so $e^{1/x} \to 0$. The limit becomes $\frac{0}{0+1} = 0$.
Step 4: Conclusion Since RHL (1) $\ne$ LHL (0), the limit does not exist. Final Answer: (D)
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