JEE PYQ: Limits Question 19
Question 19 - 2020 (08 Jan Shift 1)
$\lim_{x \to 0} \left(\frac{3x^2 + 2}{7x^2 + 2}\right)^{1/x^2}$ is equal to:
(1) $\frac{1}{e}$
(2) $\frac{1}{e^2}$
(3) $e^2$
(4) $e$
Show Answer
Answer: (2)
Solution
$L = e^{\lim_{x\to 0}\frac{1}{x^2}\ln\frac{3x^2+2}{7x^2+2}} = e^{\lim_{x\to 0}\frac{-4x^2}{x^2(7x^2+2)}} = e^{\frac{-4}{2}} = e^{-2} = \frac{1}{e^2}$.