$\lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{x^2}\right)^x=$

1

$y=\lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{x^2}\right)^x$

$\log y=\lim\limits_{x \rightarrow \infty} x \log \left(1+\frac{1}{x^2}\right)$

$=\lim\limits_{x \rightarrow \infty} x\left(\frac{1}{x^2}-\frac{1}{2 x^4}+.....\right)$

$=\lim\limits_{x \rightarrow \infty} x . \frac{1}{x^2}\left(1-\frac{1}{2 x^4}+.....\right)$

log y = 0

y = e⁰ = 1

Hence (1) is the correct answer.