Question

In Exercises \(39-42,\) use a graphing utility to (a) graph the function and (b) find the required limit (if it exists). \(\lim _{x \rightarrow \infty} \frac{x^{3}}{e^{2 x}}\)

Short answer

The limit of the function \(\frac{x^{3}}{e^{2 x}}\) as \(x\) approaches infinity is 0.

Step-by-step solution

Graph the function

The function \(\frac{x^{3}}{e^{2 x}}\) is plotted using a graphing tool such as a graphing calculator or software like Desmos or GeoGebra. The visual representation of the function gives an initial idea about its behavior as \(x\) goes towards infinity.

Apply L'Hopital's rule

The function has the form of \(\frac{\infty}{\infty}\) as \(x\) goes to infinity, which is an indeterminate form. The method to solve this is to apply L'Hopital's rule, which states that \(\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)}\), if the latter limit exists. Here, the derivative of \(x^{3}\) is \(3x^{2}\), and the derivative of \(e^{2x}\) is \(2e^{2x}\). Therefore, the function now changes to \(\frac{3x^{2}}{2e^{2x}}\).

Re-apply L'Hopital's rule

Re-apply the L'Hopital's rule on the function \(\frac{3x^{2}}{2e^{2x}}\), because that still has the form \(\frac{\infty}{\infty}\). The derivative of \(3x^{2}\) is \(6x\) and the derivative of \(2e^{2x}\) is \(4e^{2x}\). Therefore, the function now changes to \(\frac{6x}{4e^{2x}}\).

Re-apply L'Hopital's rule again

The function \(\frac{6x}{4e^{2x}}\) still has the form \(\frac{\infty}{\infty}\), thus, L'Hopital's rule is applied again. The derivative of \(6x\) is \(6\) and the derivative of \(4e^{2x}\) is \(8e^{2x}\). Therefore, the function now changes to \(\frac{6}{8e^{2x}}\).

Calculate the limit

After applying L'Hopital's rule three times, the form of the function changes from an indeterminate form to \(\frac{constant}{\infty}\), which is always zero. Therefore, the limit of the original function as \(x\) approaches infinity is zero.