Weird limit with $e^x$ 
Here's a limit that is testing my strengths.
$$\lim\limits_{x\to -\infty} [(x^2+1)e^x]$$

Personal work:
$$\lim\limits_{x\to -\infty} [(x^2+1)e^x] = \lim\limits_{x\to -\infty} (x^2e^x+e^x) = \lim\limits_{x\to -\infty} (x^2e^x)=L.$$
Let $x^2=u \iff x=-\sqrt u$, then $u_0=\lim\limits_{x\to+\infty}{x^2}=+\infty$
So, $$ L=\lim\limits_{u\to+\infty}{(u*e^{-\sqrt u})} =...$$
Although it looks correct for me, both Microsoft mathematics and symbolab show me the answer "$0$" so what am I doing wrong?
 A: HINT
Let $y=-x\to \infty$, then
$$\lim\limits_{x\to -\infty} [(x^2+1)e^x]=\lim\limits_{y\to \infty} [(y^2+1)e^{-y}]=\lim\limits_{y\to \infty} \frac{y^2+1}{e^{y}}$$
A: Write your limit in the form $$\frac{x^2+1}{e^{-x}}$$ and the searched limit is zero
A: A heuristic approach is to keep in mind that at $\pm\infty$ the exponential is stronger than any polynomial. This means that for any polynomial $p(x)$ we have the two limits
$$\lim_{x\to +\infty}{e^x\over p(x)}=+\infty$$
$$\lim_{x\to -\infty}p(x)e^x=0$$
Our problem is in the second case. Let’s prove it. Can you convince yourself that if we prove
$$\forall m\geq 0 \lim_{x\to-\infty} x^me^x=0$$
we are done. (Yes indeed a polynomial is just a finite sum of monomials).
Let’s set $u=e^x$ this means $x=\ln{u}$ and $u\to 0$. Our limit rewrites
$$\lim_{u\to 0}mu\ln{u}=m\cdot \lim_{u\to 0}u\ln{u}=0$$
A: $$\lim_{x\to -\infty} (x^2+1)e^x=\lim_{x\to \infty} \frac {x^2+1}{e^x}$$
Now by L'Hospital we have $$\lim_{x\to \infty} \frac {x^2+1}{e^x}=\lim_{x\to \infty} \frac {2x}{e^x}=\lim_{x\to \infty} \frac {2}{e^x}=0$$
