Question

lim x tends to 0 (e^2x -1)/x​

Answer 1

Step-by-step explanation:

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Answer 2

The value of $\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$ is 2.

Step-by-step explanation:

Given:

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$

To Find:

The value of $\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$ .

Solution:

As given, $\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$.

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$

$=\frac{e^{2\times0}-1 }{0} =\frac{1-1}{0} =\frac{0}{0}$

The limit of numerator and denominator is form of $\frac{0}{0}$.

Therefore, applying the  L' Hospital's Rule.

$\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}=\lim_{x \to \ 0} \frac{\frac{d}{dx} (e^{2x}-1) }{\frac{x}{dx} (x)}$

                     $=\lim_{x \to \ 0} \frac{(2e^{2x}-0 )}{1}$

                     $=\lim_{x \to \ 0} 2e^{2x}$

                      $=2\lim_{x \to \ 0} e^{2x}$

                      $=2 e^{2\times0}$

                      $=2 e^{0}\\=2\times1=2$

Thus,the value of $\lim_{x \to \ 0} \frac{e^{2x}-1 }{x}$ is 2.

                     

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