Hey,
Can anyone please tell me why is this limit correct?
$\displaystyle \lim_{x\to 0}\frac{e^{\frac{-1}{x^{2}}}}{x}$
Let $\displaystyle t=\frac{1}{x^{2}}, \;\ x=\frac{1}{\sqrt{t}}$
Making the subs gives us:
$\displaystyle \lim_{t\to \infty}\frac{\sqrt{t}}{e^{t}}=0$
e grows much more than the radical, so it tends to 0