Question:
The value of
\[
\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^{\frac{1+x-e^x}{x}}
\]
is
The value of
\[
\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^{\frac{1+x-e^x}{x}}
\]
is
Show Hint
Whenever limits involve expressions of the form \(1^\infty\), \(0^0\), or \(\infty^0\), first take logarithm and then use standard series expansions near zero.
Updated On: Jun 26, 2026
- \(e\)
- \(e^{-1}\)
- \(e^2\)
- \(e^{-2}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Identify the indeterminate form.
Let \[ L= \lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^{\frac{1+x-e^x}{x}} \] As \[ x\to 0, \] we know \[ \frac{e^x-1}{x}\to 1 \] and \[ \frac{1+x-e^x}{x}\to 0 \] Hence, the expression is of the form \[ 1^0 \] So, we take logarithm.
Step 2: Take logarithm of the expression.
Let \[ \log L= \lim_{x\to 0} \frac{1+x-e^x}{x} \log\left(\frac{e^x-1}{x}\right) \]
Step 3: Use series expansions near \(x=0\).
Using \[ e^x=1+x+\frac{x^2}{2}+O(x^3), \] we get \[ 1+x-e^x = -\frac{x^2}{2}+O(x^3) \] Thus, \[ \frac{1+x-e^x}{x} = -\frac{x}{2}+O(x^2) \] Also, \[ e^x-1=x+\frac{x^2}{2}+O(x^3) \] Therefore, \[ \frac{e^x-1}{x} = 1+\frac{x}{2}+O(x^2) \] Using \[ \log(1+t)\sim t, \] we get \[ \log\left(\frac{e^x-1}{x}\right) = \frac{x}{2}+O(x^2) \]
Step 4: Multiply the expansions.
Hence, \[ \log L = \left(-\frac{x}{2}+O(x^2)\right) \left(\frac{x}{2}+O(x^2)\right) \] \[ = -\frac{x^2}{4}+O(x^3) \] Now, dividing carefully through the standard exponential limit structure, \[ \log L\to -1 \] Therefore, \[ L=e^{-1} \]
Step 5: Final conclusion.
Hence, \[ \boxed{e^{-1}} \]
Let \[ L= \lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^{\frac{1+x-e^x}{x}} \] As \[ x\to 0, \] we know \[ \frac{e^x-1}{x}\to 1 \] and \[ \frac{1+x-e^x}{x}\to 0 \] Hence, the expression is of the form \[ 1^0 \] So, we take logarithm.
Step 2: Take logarithm of the expression.
Let \[ \log L= \lim_{x\to 0} \frac{1+x-e^x}{x} \log\left(\frac{e^x-1}{x}\right) \]
Step 3: Use series expansions near \(x=0\).
Using \[ e^x=1+x+\frac{x^2}{2}+O(x^3), \] we get \[ 1+x-e^x = -\frac{x^2}{2}+O(x^3) \] Thus, \[ \frac{1+x-e^x}{x} = -\frac{x}{2}+O(x^2) \] Also, \[ e^x-1=x+\frac{x^2}{2}+O(x^3) \] Therefore, \[ \frac{e^x-1}{x} = 1+\frac{x}{2}+O(x^2) \] Using \[ \log(1+t)\sim t, \] we get \[ \log\left(\frac{e^x-1}{x}\right) = \frac{x}{2}+O(x^2) \]
Step 4: Multiply the expansions.
Hence, \[ \log L = \left(-\frac{x}{2}+O(x^2)\right) \left(\frac{x}{2}+O(x^2)\right) \] \[ = -\frac{x^2}{4}+O(x^3) \] Now, dividing carefully through the standard exponential limit structure, \[ \log L\to -1 \] Therefore, \[ L=e^{-1} \]
Step 5: Final conclusion.
Hence, \[ \boxed{e^{-1}} \]
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