Evaluate the limit of x^3 * e^(-2x) as x approaches infinity.

Step-by-step solution

1. 1

   Turn $\infty\cdot 0$ into $\infty/\infty$. As written the limit is a product of $x^3\to\infty$ and $e^{-2x}\to 0$, a form no rule applies to directly. Moving the exponential to the denominator fixes that:

   $x^{3}e^{-2x}=\frac{x^{3}}{e^{2x}}$

   Now both parts tend to $\infty$, which is a form L Hopital handles.

2. 2

   Differentiate top and bottom once.

   $\lim_{x\to\infty}\frac{x^{3}}{e^{2x}}=\lim_{x\to\infty}\frac{3x^{2}}{2e^{2x}}$

   Each pass lowers the numerator degree by one but only multiplies the denominator by a constant $2$ — the exponential never gets smaller.

3. 3

   Differentiate twice more.

   $\lim_{x\to\infty}\frac{3x^{2}}{2e^{2x}}=\lim_{x\to\infty}\frac{6x}{4e^{2x}}=\lim_{x\to\infty}\frac{6}{8e^{2x}}$

   Each step is still $\infty/\infty$ until the numerator becomes a constant, which is when to stop.

4. 4

   Evaluate the final expression. The numerator is now fixed at $6$ while $e^{2x}\to\infty$, so

   $\frac{6}{8e^{2x}}\to 0\ \Longrightarrow\ \lim_{x\to\infty}x^{3}e^{-2x}=0$

5. 5

   Recognise the general principle. For any $n>0$ and $a>0$, $\displaystyle\lim_{x\to\infty}x^{n}e^{-ax}=0$: exponentials beat every polynomial. Knowing this saves the three L Hopital passes on sight. Numerically $x^3e^{-2x}$ is $3.4\times 10^{-14}$ at $x=20$ and $4.7\times 10^{-39}$ at $x=50$ ✓.