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I know that $\lim \frac{x^n}{e^x} = 0$ because $e^x$ grows faster than $x^n$.

So, in this case, I think that

$$\lim_{x \rightarrow - \infty}(x^9e^{2x}) = 0$$ because $e^x$ is going to tend to zero faster than $x^9$ is going to tend to $- \infty$. (does that make sense?)

Is that correct? How do I solve this?

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    Take a look at http://math.stackexchange.com/questions/55468/how-to-prove-that-exponential-grows-faster-than-polynomial . There are proofs without L'Hopital, – 2017-02-23

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Let $y = -2x$

$$\lim_{x \rightarrow - \infty } x^9 e^{2x}=\lim_{y \rightarrow \infty}\frac{y^9}{(-2)^9e^y}=\frac{1}{(-2)^9}\lim_{y \rightarrow \infty}\frac{y^9}{e^y}=0$$

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    that y^9/e^y stuff seems intuitive (and pretty obvious) but how could you PROVE it? – 2017-02-23
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    The OP already knows that from the first sentence isn't it? – 2017-02-23
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    How does $\lim_{x \rightarrow - \infty } x^9 e^{2x}=\lim_{y \rightarrow \infty}\frac{y^9}{(-2)^9e^y}$ ? What did you do? – 2017-02-23
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    I substitute $y = -2x$ , hence we have $y^9 = (-2x)^9$ and $2x= -y$. Also as $x \rightarrow - \infty$, $y \rightarrow \infty$. – 2017-02-23
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I found a simpler way to solve this:

$$\lim_{x \rightarrow - \infty}(x^9e^{2x}) = x^9 \cdot \frac{1}{e^{-2x}} = \frac{x^9}{e^{-2x}}$$

And since $\lim_{x \rightarrow -\infty} -2x = \infty$ and $\lim \frac{x^n}{e^x} = 0$, I followed Siong Thye Goh's advice and made $-2x = y$:

$$\frac{x^9}{e^{-2x}} = \frac{(\frac{y}{-2})^9}{e^y} = \frac{1}{(-2)^9} \cdot \frac{y^9}{e^y} = \frac{1}{(-2)^9} \cdot 0 = 0$$

Did I make any mistake?

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You seem concerned about people relying on the growth property of the exponential function without proving it. I'll give a proof of both, which should hopefully scratch that mental itch.

As a preliminary result, it is possible to derive the exponential power series by elementary means (i.e., excluding derivative methods, Taylor series, l'Hopital's rule, etc), as given, for instance, in equivalence of exponential definitions. This shows $$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

Obviously if $x>0$ then all of the terms of this series are positive, so choosing just one of them must be less than the entire series, that is $$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \geq \frac{x^n}{n!},\quad\text{for any $n=1,2,\dotsc$}$$

Now, applying this idea with $n=k+9$, $$ \frac{x^9}{e^{2x}} \leq \frac{x^9}{(2x)^{k+9}/(k+9)!} = \frac{(k+9)!}{2^{k+9}} \cdot \frac{1}{x^k} $$ If $k > 0$, then $$ 0\leq \lim_{x\to\infty} \frac{x^9}{e^{2x}} \leq \frac{(k+9)!}{2^{k+9}}\lim_{x\to\infty} \frac{1}{x^k} = 0$$

So for the original problem, $$ \lim_{x\to-\infty} x^9 e^{2x} = \lim_{y = -x \to\infty} (-y)^9 e^{2(-y)} = - \lim_{y\to\infty} \frac{y^9}{e^{2y}} = 0$$

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Consider the statement $\int_0^\infty x^n e^{-x}dx=n!,\,\lim_{x\to\infty}x^n e^{-x}=0$. If the first half is true for a given $n\ge 0$, the second half is true too as otherwise the integral would diverge. The full statement can therefore be proven by induction, with the second half of the $n=k$ case proving the first half of the $n=k+1$ case by integration by parts.