Prove that $f(x)=e^{-1/x^2}$ for $x\neq 0$ is continuous. 
Prove that $f(x)=e^{-1/x^2}$ for $x\neq 0$ is continuous.

So I know I need to use the epsilon-delta limit definition, but when I do I end up needing to show that
$e^{-1/a^2}\left|\dfrac{2}{x^3}e^{-1/x^2+1/a^2}-\dfrac{2}{a^3}\right|<\epsilon\; \forall \epsilon>0.$
I can simplify it to $e^{-1/a^2}\left|\dfrac{2}{x^3}e^{-(x^2-a^2)/(x^2a^2)}-\dfrac{2}{a^3}\right|.$ Then if I let $\delta <1,$ I get $x>a-1$ and so the expression in the absolute value brackets is less than $\dfrac{2}{(a-1)^3}e^{-(x^2-a^2)/((a-1)^2a^2)}-\dfrac{2}{a^3},$ but I don't know how to simplify this to get a fraction. I don't want to use the Taylor expansion for $e.$
Any help would be appreciated.

We may evaluate the continuity of $e^{\frac{-1}{x^2}}$ straightforwardly using theorems regarding the continuity of composed and elementary functions.
Let $f(x)=\frac{1}{x},$ $g(x) = e^x$, and $h(x)=\frac{1}{x^2}$.
Then $$e^{\frac{-1}{x^2}} = \frac{1}{e^{\frac{1}{x^2}}} = \frac{1}{(g \circ h)(x)}= (f \circ (g \circ h))(x)$$.
We know $h(x)=\frac{1}{x^2}$ is a rational function, so it is continuous on its domain, $\mathbb{R}-\{0\}$. We know $g(x) = e^x$ is an exponential function, so it is continuous on its domain, $\mathbb{R}$. Therefore, $(g \circ h) (x) = e^{\frac{1}{x^2}}$ is continuous on the intersection of the two domains, which is $\mathbb{R}-\{0\}$. 
Similar to before, we know $f(x)=\frac{1}{x}$ is a rational function, so it is continuous on its domain, $\mathbb{R}-\{0\}$. And we have already stated that $(g \circ h) (x) = e^{\frac{1}{x^2}}$ is continuous on its domain, $\mathbb{R}-\{0\}$. So $(f \circ (g \circ h))(x)$ is continuous on the intersection of the two domains, which is $\mathbb{R}-\{0\}$.
Therefore, $e^{\frac{-1}{x^2}}=(f \circ (g \circ h))(x)$ is continuous on $\mathbb{R}-\{0\}$.
