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Q. 53

Expert-verified
Found in: Page 136

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# $\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)$

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2$

See the step by step solution

## Step 1. Given information

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)$

## Step 2.Simplify

Since,

$\underset{x\to a}{\mathrm{lim}}\left[c·f\left(x\right)\right]=c·\underset{x\to a}{\mathrm{lim}}f\left(x\right)$

Therefore,

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2·\underset{x\to 0}{\mathrm{lim}}\left(\frac{x}{{e}^{x}-1}\right)$

## Step 3. Calculate the value

Apply L'Hopital's rule,

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2·\underset{x\to 0}{\mathrm{lim}}\left(\frac{1}{{e}^{x}}\right)$

role="math" localid="1648011355231" $\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2\frac{1}{{e}^{0}}$

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2$