## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}(x^3e^{-x^2})=0$$
$$A=\lim_{x\to\infty}(x^3e^{-x^2})$$ $$A=\lim_{x\to\infty}\frac{x^3}{e^{x^2}}$$ As $x\to\infty$, both $x^3$ and $e^{x^2}$ would approach $\infty$. So we have an indeterminate form of $\infty/\infty$. With the use of L'Hospital's Rule: $$A=\lim_{x\to\infty}\frac{(x^3)'}{(e^{x^2})'}$$ $$A=\lim_{x\to\infty}\frac{3x^2}{2xe^{x^2}}$$ $$A=\lim_{x\to\infty}\frac{3x}{2e^{x^2}}$$ As $x\to\infty$, both $3x$ and $2e^{x^2}$ approach $\infty$. We encounter another indeterminate form of $\infty/\infty$ and apply L'Hospital's Rule here: $$A=\lim_{x\to\infty}\frac{(3x)'}{(2e^{x^2})'}$$ $$A=\lim_{x\to\infty}\frac{3}{4xe^{x^2}}$$ As $x\to\infty$, $4xe^{x^2}\to\infty$. Therefore, $\frac{3}{4xe^{x^2}}\to0$. In other words, $$A=0$$