MereTeacher

JEE • NEET • FOUNDATION

JEE Previous Year Question

$lim x \to 0 (\frac{3 x ^{2} + 2}{7 x ^{2} + 2})^{\frac{1}{x ^{2}}} is equal to:$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $e^{- 2}$
B $e^{2}$
C $e^{2} 7$
D $e^{\frac{3}{7}}$

Correct answer: A

Solution

(a) JEE Main 2020 Let L = lim x \to 0 (\frac{3 x ^{2} + 2}{7 x ^{2} + 2})^{1/ x^{2}} = e^{lim x \to 0 \frac{1}{x ^{2}} (\frac{3 x ^{2} + 2}{7 x ^{2} + 2} - 1)} = e^{lim x \to 0 \frac{1}{x ^{2}} (\frac{- 4 x ^{2}}{7 x ^{2} + 2})} = e^{- \frac{4}{2}} = e^{- 2}

More Limits PYQs