lim-x-1-1-x-x-2-e-x-

Question Number 212798 by MrGaster last updated on 24/Oct/24

\lim_{x \to \infty} {(1 + \frac{1}{x})}^{\frac{x^{2}}{e^{x}}} = ?

Answered by mehdee7396 last updated on 24/Oct/24

{l i m}_{x \to \infty} (1 + \frac{1}{x}) = 1 & {l i m}_{x \to \infty} \frac{x^{2}}{e^{x}} = 0

A = {(1 + \frac{1}{x})}^{\frac{x^{2}}{e^{x}}}

\Rightarrow {l i m}_{x \to \infty} A = {l i m}_{x \to \infty} \frac{x^{2}}{e^{x}} l n (1 + \frac{1}{x})

= 0 \times 0 = 0 \Rightarrow A = 1