I have used L'Hospital's Rule on $\lim_{x \to \infty} x^3e^{-x^2}$ to get $\lim_{x \to \infty} 3x^2e^{-x^2}-2x^4e^{-x^2}$. I plugged the latter form of the limit into desmos.com and found that as $x$ approaches infinity, the limit equals zero.
$$\lim_{x \to \infty} x^3e^{-x^2} = \lim_{x \to \infty} \frac{d}{dx} x^3e^{-x^2} = \lim_{x \to \infty} [3x^2 \cdot e^{-x^2}]+[x^3 \cdot -2xe^{-x^2}] = \lim_{x \to \infty}3x^2e^{-x^2}-2x^4e^{-x^2}$$
However, I can't rely on desmos in an exam, and plugging in multiple values for x takes a long time. Is there an easier way to solve limits to infinity other than plugging in multiple values for $x$?
Best Answer
Hint: The limit is equal to
$$\lim\limits_{x\to\infty} \frac{x^3}{e^{x^2}} \stackrel{\text{L'Hospital}}{=} \lim\limits_{x\to\infty}\frac{3x^2}{2xe^{x^2}} \stackrel{\text{L'Hospital}}{=} \cdots$$