Symmetry of m(x)

Compute $m(-x) = (-x)^2 e^{-(-x)^2} = x^2 e^{-x^2}$. This is exactly equal to $m(x)$, since both $(-x)^2 = x^2$ and $(-x)^2 = x^2$ inside the exponential.

Since $m(-x) = m(x)$ for all $x$, the function satisfies the definition of an even function. Therefore $m(x)$ is symmetric about the $y$-axis.

By the product rule, $m'(x) = 2xe^{-x^2} + x^2(-2x)e^{-x^2} = 2xe^{-x^2}(1 - x^2)$. Setting $m'(x) = 0$ gives $x = 0, 1, -1$ since $e^{-x^2}$ is never zero.

Testing signs of $m'(x) = 2xe^{-x^2}(1-x^2)$ around these points: for $x$ slightly less than $-1$, $m'(x)>0$; between $-1$ and $0$, $m'(x)<0$; between $0$ and $1$, $m'(x)>0$; and beyond $1$, $m'(x)<0$.

This sign pattern shows $x=-1$ is a local maximum, $x=0$ is a local minimum, and $x=1$ is a local maximum. Evaluating $m$ at these points gives $m(-1)=m(1)=e^{-1}$ and $m(0)=0$.

Rewrite $m(x) = x^2 e^{-x^2} = \dfrac{x^2}{e^{x^2}}$. As $x \to \infty$, numerator and denominator both tend to $\infty$, so this is an $\infty/\infty$ indeterminate form, justifying L'Hopital's Rule.

Differentiating numerator and denominator separately: $\dfrac{d}{dx}[x^2] = 2x$ and $\dfrac{d}{dx}[e^{x^2}] = 2xe^{x^2}$, giving $\lim_{x\to\infty} \dfrac{2x}{2xe^{x^2}} = \lim_{x\to\infty} \dfrac{1}{e^{x^2}}$.

As $x \to \infty$, $e^{x^2} \to \infty$, so $\dfrac{1}{e^{x^2}} \to 0$. Thus $\lim_{x\to\infty} m(x) = 0$, and since $m$ is even, the same limit holds as $x \to -\infty$, giving a horizontal asymptote at $y=0$.