Question

Evaluate $\underset{x\to 0}{\lim }{\left(\frac{3x^2+2}{7x^2+2}\right)}^{\frac{1}{x^2}}$

• A. $e$
• B. $\frac{1}{e^2}$
• C. $\frac{1}{e}$
• D. $e^2$

Answer 1

The correct option is B

$\frac{1}{e^2}$

Explanation for correct answer:

Evaluating the given function:

Given,

$$\begin{gathered} \underset{x\to 0}{\lim }{\left(\frac{3x^2+2}{7x^2+2}\right)}^{\frac{1}{x^2}}={\left(\frac{3.0^2+2}{7.0^2+2}\right)}^{\frac{1}{0^2}} \\ ={\left(\frac{2}{2}\right)}^{\infty } \\ ={\left(1\right)}^{\infty } \end{gathered}$$

We know that,

If, $\underset{x\to 0}{\lim }f{\left(x\right)}^{g\left(x\right)}=1^{\infty },$ then

$\underset{x\to 0}{\lim }f{\left(x\right)}^{g\left(x\right)}=e^{\underset{x\to 0}{\lim }g\left(x\right).\left(f\left(x\right)-1\right)}$

Apply this formula, and we get

$\underset{x\to 0}{\lim }{\left(\frac{3x^2+2}{7x^2+2}\right)}^{\frac{1}{x^2}}=e^{\underset{x\to 0}{\lim }\frac{1}{x^2}.\left(\left(\frac{3x^2+2}{7x^2+2}\right)-1\right)}\to \left(i\right)$

First, we solve

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{1}{x^2}.\left(\left(\frac{3x^2+2}{7x^2+2}\right)-1\right)=\underset{x\to 0}{\lim }\frac{1}{x^2}.\left(\frac{3x^2+2-7x^2-2}{7x^2+2}\right) \\ =\underset{x\to 0}{\lim }\frac{\cancel{x^2}}{\cancel{x^2}}.\left(\frac{-4}{7x^2+2}\right) \\ =\left(\frac{-4}{7.0+2}\right) \\ =-\frac{4}{2} \\ =-2 \end{gathered}$$

Substituting this in equation $\left(i\right)$

$$\begin{gathered} \underset{x\to 0}{\lim }{\left(\frac{3x^2+2}{7x^2+2}\right)}^{\frac{1}{x^2}}=e^{\underset{x\to 0}{\lim }\frac{1}{x^2}.\left(\left(\frac{3x^2+2}{7x^2+2}\right)-1\right)} \\ =e^{-2} \\ =\frac{1}{e^2} \end{gathered}$$

Hence, the correct answer is option (B).