Question

Evaluate : $\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{x\left(e^{5x}-1\right)}$

• A. $\frac{3}{5}$
• B. $\frac{5}{3}$
• C. $-\frac{3}{5}$
• D. $-\frac{5}{3}$
• E. $1$

Answer 1

The correct option is A

$\frac{3}{5}$

Explanation for the correct option:

Find the value of $\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{x\left(e^{5x}-1\right)}$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{x\left(e^{5x}-1\right)} \\ I=\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{3x^2}\times \frac{5x}{e^{5x}-1}\times \frac{3}{5} \\ I=\frac{3}{5}\left(\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{3x^2}\times \underset{x\to 0}{\lim }\frac{5x}{e^{5x}-1}\right) \end{gathered}$$

We know that,

$$\begin{gathered} \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{log}\left(1+x\right)}{\mathbf{x}}\mathbf{=}\mathbf{1} \\ \underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{and}\mathbf{}\mathbf{lim}}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{1} \end{gathered}$$

Then,

$$\begin{gathered} I=\frac{3}{5}\times 1\times 1 \\ I=\frac{3}{5} \end{gathered}$$

Since,

$\underset{x\to 0}{\lim }\frac{\log \left(1+3x^2\right)}{x\left(e^{5x}-1\right)}=\frac{3}{5}$

Hence, the correct answer is option A.