Show that $\lim_{x\to 0}x^{-n}e^{-1/x^2}=0$ for any positive integer $n$ Please check if my answer is correct or not.
First, rewrite the limit as
$$
\lim_{x\to 0}x^{-n}e^{-1/x^2}=\lim_{x\to 0}\frac{x^{-n}}{e^{1/x^2}}\tag{1}
$$
Notice that limit in the right side of $(1)$ is indeterminate form $\infty/\infty$, so we can apply L'Hopital's rule
$$
\begin{align}
\lim_{x\to 0}\frac{x^{-n}}{e^{1/x^2}}&=\lim_{x\to 0}\frac{-n·x^{-n-1}}{e^{1/x^2}·(-2/x^3)}\\
&=\lim_{x\to 0}\frac{n·x^{-n+2}}{2·e^{1/x^2}}\\
&=\frac{n}{2}·\lim_{x\to 0}\frac{x^{-n+2}}{e^{1/x^2}}\tag{2}\\
\end{align}
$$
If $-n+2\geq 0$, then
$$\lim_{x\to 0}x^{-n}e^{-1/x^2}=\lim_{x\to 0}\frac{n·x^{-n+2}}{2·e^{1/x^2}}=0$$
if $-n+2<0$, $(2)$ is again indeterminate form $\infty/\infty$, then we apply L'Hopital's rule again
$$
\begin{align}
\lim_{x\to 0}\frac{n·x^{-n+2}}{2·e^{1/x^2}}&=\lim_{x\to 0}\frac{n(-n+2)x^{-n+1}}{2·e^{1/x^2}(-2/x^3)}\\
&=\lim_{x\to 0}\frac{n(2-n)x^{-n+4}}{2^2·e^{1/x^2}}\\
&=\frac{n(2-n)}{2^2}·\lim_{x\to 0}\frac{x^{-n+4}}{e^{1/x^2}}\tag{3}\\
\end{align}
$$
Again we check the sign of $-n+4$, if $-n+4\geq 0$, the limit equals zero, if not, apply L'Hopital's rule continually.
Notice that everytime we apply L'Hopital's rule, the power of limit's numerator $x$ increases by 2, given $n$ is a positive integer, if we use L'Hopital's rule continually, the power will becomes greater or equal to $0$ eventually, which implies that the limit equals $0$.
 A: Since
\begin{equation*}
x^{-n}e^{-1/x^{2}}=\frac{1}{x^{n}e^{1/x^{2}}}\text{,}
\end{equation*}
to show
\begin{equation*}
\lim_{x\rightarrow 0}\left( x^{-n}e^{-1/x^{2}}\right) =0\text{ for all }n\in 
\mathbb{Z}^{+}
\end{equation*}
amounts to
\begin{equation*}
\lim_{x\rightarrow 0}\left \vert x^{n}e^{1/x^{2}}\right \vert =\infty \text{,
where }\left \vert .\right \vert \text{ is the absolute value operator.}
\end{equation*}
Now, using Taylor Expansion around $x=0$, we have
\begin{equation*}
e^{1/x^{2}}=\sum_{k=0}^{\infty }\frac{1}{k!\times x^{2k}}\text{,}
\end{equation*}
which implies
\begin{equation*}
\lim_{x\rightarrow 0}\left \vert x^{n}e^{1/x^{2}}\right \vert
=\lim_{x\rightarrow 0}\left( \sum_{k=0}^{\infty }\left \vert \frac{x^{n-2k}}{
k!}\right \vert \right) =\sum_{k=0}^{\infty }\lim_{x\rightarrow 0}\left \vert 
\frac{x^{n-2k}}{k!}\right \vert \text{.}
\end{equation*}
Finally, notice that
\begin{equation*}
\lim_{x\rightarrow 0}\left \vert \frac{x^{n-2k}}{k!}\right \vert =\left \{ 
\begin{array}{cc}
\infty  & \text{if }k>n/2 \\ 
1/k! & \text{if }k=n/2 \\ 
0 & \text{if }k<n/2
\end{array}
\right. \text{.}
\end{equation*}
Hence,
\begin{equation*}
\lim_{x\rightarrow 0}\left \vert x^{n}e^{1/x^{2}}\right \vert =\infty \text{,}
\end{equation*}
which, as stated at the beginning, suffices to prove%
\begin{equation*}
\lim_{x\rightarrow 0}\left( x^{-n}e^{-1/x^{2}}\right) =0\text{.}
\end{equation*}
