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  • Evaluate    <img alt="\quad \lim _{x \rightarrow 0} \frac{x \cdot \ln (1+x)}{e^{2 x}-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cquad%20%5Clim%20_%7Bx%20%5Crightarrow

Evaluate    \quad \lim _{x \rightarrow 0} \frac{x \cdot \ln (1+x)}{e^{2 x}-1}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Using L-hospital

\lim _{x \rightarrow 0} \frac{x \cdot \ln (1+x)}{e^{2 x}-1}

Now as ! approaches to 0 it is in \frac{0}{0} so just apply L'Hospital's rule differentiating numerator and denominator we get

\lim _{x \rightarrow 0} \frac{\ln (1+x)+\frac{x}{1+x}}{e^{2 x} \times 2}

Which is  0 / 1=0

Posted by

Irshad Anwar

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