Evaluate $\displaystyle\lim_{x\to \infty} \left(1+\dfrac{1}{x^2} \right)^{x}$ without l'hôpital.
What I have tried:
$\displaystyle\lim_{x\to \infty} \left(1+\dfrac{1}{x^2} \right)^{x} = \displaystyle\lim_{x\to \infty}\left[ \left(1+\dfrac{1}{x^2} \right)^{x^2}\right]^{\frac{1}{x}}$
I would really like to say that is $e^0$, but that's probably illegal. Is there another way?