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

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

Question
Limits and continuity
asked 2021-03-04
Find the following limit:
\(\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}\)

Answers (1)

2021-03-05
Given:
\(\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}\)
Now Applying limits:
\(\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}=\frac{0(e^0)}{e^{3(0)}-1}=\frac{0}{0}\)
Hence it gives indeterminate form \(\frac{0}{0}\)
Now by L-Hospital Rule:
\(\lim_{x\rightarrow0}[\frac{\frac{d}{dx}(xe^x)}{\frac{d}{dx}(e^{3x}-1)}]=\lim_{x\rightarrow0}[\frac{xe^x+e^x}{3e^{3x}}]\)
\(=[\frac{0(e^0)+e^0}{3e^{3(0)}}]\)
\([=\frac{1}{3}]\)
Hence,
\(\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}=\frac{1}{3}\)
0

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