i've been unsuccessful with attempting this epsilon delta proof.
$\lim_{x \rightarrow 0}e^\frac{-1}{x^2}=0$
So far i've been attempting to utilize the following inequalities for an upper bound from my textbook:
$1+x\leq e^x \leq \frac{1}{1-x}$ and the fact that $a \leq b \leq c$ $\rightarrow$ $\lvert b\rvert \leq \lvert a\rvert + \lvert c\rvert$.
i then get the following:
$\lvert e^\frac{-1}{x^2} \rvert \leq \lvert \dfrac{1}{1+\frac{1}{x^2}} \rvert + \lvert - \dfrac{1}{x^2} \rvert+1$
But i can't seem to move on from here. Thanks in advance.
We want to show that for all $\epsilon > 0$, there exists $\delta > 0$ s.t. $$|x| < \delta \implies \big|e^\frac{-1}{x^2}\big| < \epsilon$$
Suppose $|x| < \delta$,
$$\begin{align} |x| < \delta &\implies x^2 < \delta^2 \\ &\implies \frac{1}{x^2} > \frac{1}{\delta^2} \\ &\implies \frac{-1}{x^2} < \frac{-1}{\delta^2} \\ \end{align}$$
and since $e^x$ is a monotonically increasing function the result above implies that
$$e^\frac{-1}{x^2} < e^\frac{-1}{\delta^2}$$
and since $e^x$ is always positive we can see that
$$\big|e^\frac{-1}{x^2}\big| < e^\frac{-1}{\delta^2}$$
now determining $\delta$ is just a matter of solving the equation
$$e^\frac{-1}{\delta^2} = \epsilon$$
which gives us
$$\delta = {1 \over \sqrt{-\ln \epsilon}}$$
you can also see that this function for $\delta$ is not defined at $\epsilon \ge 1$ so we need to find a $\delta$ for cases where $\epsilon \ge 1$ which turns out to be trivial since the function $e^\frac{-1}{x^2}$ is always less than $1$ so any positive value for $\delta$ would work, we can choose an arbitrary value like $1$ for example.
Thus our $\delta$ can be
$$\begin{align} \delta = \begin{cases} {1 \over \sqrt{-\ln \epsilon}} & \epsilon < 1 \\ 1 &\epsilon \ge 1 \\ \end{cases} \end{align}$$