Expert Answer to "Evaluate limx→0+1x−1ex−1"

Bergen

Answered question

2021-10-28

Evaluate

lim_{x \to 0^{+}} \frac{1}{x} - \frac{1}{e^{x} - 1}

Arham Warner

Skilled2021-10-29Added 102 answers

L-Hospitals Rule:
If

lim_{x \to a} \frac{f (a)}{g (a)} = \frac{0}{0}

, then

lim_{x \to a} \frac{f (a)}{g (a)} = lim_{x \to a} \frac{\frac{d}{d x} [f (a)]}{\frac{d}{d x} [g (a)]}

To evaluate:

lim_{x \to 0^{+}} \frac{1}{x} - \frac{1}{e^{x} - 1}

= lim_{x \to 0^{+}} \frac{e^{x} - 1 - x}{x (e^{x} - 1)} = lim_{x \to 0^{+}} \frac{e^{x} - x - 1}{x e^{x} - x}

x \to 0^{+}, lim_{x \to 0} \frac{e^{x} - x - 1}{x e^{x} - x} = \frac{0}{0}

indeterminate
As

x \to 0^{+}, lim_{x \to 0^{+}} \frac{e^{x} - x - 1}{x e^{x} - x} = \frac{e^{0} - 0 - 1}{0 \times e^{0} - 0} = \frac{1 - 1}{0} = \frac{0}{0}

which is intederminate form we can use L-Hospitals Rule:

∴ lim_{x \to 0^{+}} \frac{e^{x} - x - 1}{x e^{x} - x} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} (e^{x} - x - 1)}{\frac{d}{d x} (x e^{x} - x)}

= lim_{x \to 0^{+}} \frac{e^{x} - 1 - 0}{x e^{x} + e^{x} - 1} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} (e^{x} - 1)}{\frac{d}{d x} (x e^{x} + e^{x} - 1)}

This is helpful 0 Flag

Jeffrey Jordon

Expert2022-08-30Added 2605 answers

Answer is given below (on video)

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