Math / Calculus

Question17. [-/1 Points] DETAILS MY NOTES SCALCET9M 4.4.053.
Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. $\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$ $\square$ Need Help? Read It

Studdy Solution
Evaluate the limit:
As $x \to 0^+$ , the expression simplifies to:
$\frac{e^0 - 1}{e^0 - 1 + 0 \cdot e^0} = \frac{0}{0} + 0 = 0$
Thus, the limit is:
$\lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x} = 0$
The limit is $0$ .

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