Find the following limit: lim_{xrightarrow0}frac{xe^x}{e^{3x}-1}

Find the following limit:
$\underset{x\to 0}{lim}\frac{x{e}^{x}}{{e}^{3x}-1}$
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casincal
Given:
$\underset{x\to 0}{lim}\frac{x{e}^{x}}{{e}^{3x}-1}$
Now Applying limits:
$\underset{x\to 0}{lim}\frac{x{e}^{x}}{{e}^{3x}-1}=\frac{0\left({e}^{0}\right)}{{e}^{3\left(0\right)}-1}=\frac{0}{0}$
Hence it gives indeterminate form $\frac{0}{0}$
Now by L-Hospital Rule:
$\underset{x\to 0}{lim}\left[\frac{\frac{d}{dx}\left(x{e}^{x}\right)}{\frac{d}{dx}\left({e}^{3x}-1\right)}\right]=\underset{x\to 0}{lim}\left[\frac{x{e}^{x}+{e}^{x}}{3{e}^{3x}}\right]$
$=\left[\frac{0\left({e}^{0}\right)+{e}^{0}}{3{e}^{3\left(0\right)}}\right]$
$\left[=\frac{1}{3}\right]$
Hence,
$\underset{x\to 0}{lim}\frac{x{e}^{x}}{{e}^{3x}-1}=\frac{1}{3}$
Jeffrey Jordon