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

Question
Limits and continuity
Find the following limit:
$$\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}$$

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,
$$\lim_{x\rightarrow0}\frac{xe^x}{e^{3x}-1}=\frac{1}{3}$$

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