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Q. 53

Expert-verifiedFound in: Page 136

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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

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

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

Since,

$\underset{x\to a}{\mathrm{lim}}\left[c\xb7f\left(x\right)\right]=c\xb7\underset{x\to a}{\mathrm{lim}}f\left(x\right)$Therefore,

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2\xb7\underset{x\to \hspace{0.17em}0}{\mathrm{lim}}\left(\frac{x}{{e}^{x}-1}\right)$Apply L'Hopital's rule,

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2x}{{e}^{x}-1}\right)=2\xb7\underset{x\to \hspace{0.17em}0}{\mathrm{lim}}\left(\frac{1}{{e}^{x}}\right)$

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

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

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