The value of lim_{x -> -inf} (1 + x^2)^(1/x) is A. 0 B. 1/2 C. 1 D. infty

GATE 2015 · Engineering Mathematics · Limits · medium

Answer: 1

1. Take logarithm: Set L = lim (1+x^2)^(1/x). Then ln L = lim_{x->-inf} ln(1+x^2) / x.
2. Evaluate the ratio: As x -> -inf, ln(1+x^2) ~ 2*ln|x|, which grows much slower than |x|. So ln(1+x^2)/|x| -> 0. Thus ln L = 0.
3. Exponentiate: Since ln L = 0, we get L = e^0 = 1.