Question

$\displaystyle\lim_{x \to 0} \left(\frac{3x^{2}+2}{7x^{2}+2}\right)^{\frac {1}{x^2}}$ is equal to :

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

Answer 1

Answer: (B)

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

Answer 2

Required limit $=e^{\displaystyle \lim_{x \to 0}\left(\frac{3x^{2}+2}{7x^{2}+2}-1\right) \frac{1}{x^{2}} }$
$=e^{\displaystyle \lim_{x \to 0}\left(\frac{-4}{7x^{2}+2}\right)}$ $=\frac{1}{e^{2}}$

The correct option is (B): $\frac {1}{e^2}$.