Evaluate the limit: _x infinity (x3)/(e^x)

As $x \to \infty$, the numerator $x^3$ approaches $\infty$, and the denominator $e^x$ also approaches $\infty$. This is an indeterminate form of type $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule.

L'Hôpital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$.
Here, $f(x) = x^3$ and $g(x) = e^x$.
The derivatives are $f'(x) = 3x^2$ and $g'(x) = e^x$.

The limit is still of the form $\frac{\infty}{\infty}$ as $x \to \infty$.
Let $f_1(x) = 3x^2$ and $g_1(x) = e^x$.
The derivatives are $f_1'(x) = 6x$ and $g_1'(x) = e^x$.

The limit is still of the form $\frac{\infty}{\infty}$ as $x \to \infty$.
Let $f_2(x) = 6x$ and $g_2(x) = e^x$.
The derivatives are $f_2'(x) = 6$ and $g_2'(x) = e^x$.

As $x \to \infty$, $e^x$ approaches $\infty$.
Therefore, $\frac{6}{e^x}$ approaches $\frac{6}{\infty}$, which is 0.