AP Calculus AB and BC

Limit of x^2e^(-x) as x Approaches Infinity Is 0

Q: Why is infinity times 0 not simply 0?

Because the factor heading to 0 may be shrinking more slowly than the other is growing. The product x^2 \cdot \frac{1}{x} has exactly that shape and runs to \infty. Only a comparison of rates settles it, and turning the product into a quotient is what makes the comparison possible.

Q: Which factor should go in the denominator?

The one that gets simpler when it moves. Sending e^{-x} down produces \frac{x^2}{e^x}, where each derivative shrinks the numerator toward a constant. Sending x^2 down produces \frac{e^{-x}}{x^{-2}}, where the derivatives grow more complicated. Both are valid; only one finishes.

Q: Does the answer change for x^{10}e^{-x}?

No, it is still 0, though it takes ten passes rather than two and the approach is slower. That function climbs to its maximum at x = 10 before turning, so a table sampled below x = 10 would suggest the opposite answer.

The limit of x squared times e to the negative x as x approaches infinity is 0. As a product it reads infinity times 0, which decides nothing, so rewrite it as x squared over e to the x. That is infinity over infinity, and two passes of L'Hopital leave 2 over e to the x, which goes to 0.

\lim_{x \to \infty} x^2e^{-x} = 0

Settled by L'Hopital's rule applied twice.

Rewriting as a quotient, then two passes

L'Hopital's rule reads quotients, not products, so the first move is to send the decaying factor downstairs using $e^{-x} = \frac{1}{e^x}$ .

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

That quotient has the form $\frac{\infty}{\infty}$ , so the rule applies. One pass leaves $\frac{2x}{e^x}$ , which is still $\frac{\infty}{\infty}$ , and that licenses a second.

\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0

Recheck the form between passes rather than at the end. The intermediate $\frac{2x}{e^x}$ is indeterminate, the final $\frac{2}{e^x}$ is not, so the work stops there with the value $0$ .

$x$	$x^2$	$x^2e^{-x}$
$10$	$100$	$0.00454$
$20$	$400$	$0.000000824$
$50$	$2500$	$4.8 \cdot 10^{-19}$

Why the product form decides nothing

Substituting directly gives $\infty \cdot 0$ : the square grows without bound while the exponential decays to nothing. That combination is indeterminate, because the outcome depends on which factor moves faster.

Two quick examples make the point. The product $x^2 \cdot \frac{1}{x}$ has the same $\infty \cdot 0$ shape and runs to $\infty$ , while $x \cdot \frac{1}{x^2}$ has it too and collapses to $0$ . The shape alone predicts neither.

Send the exponential down, not the polynomial

Rewriting as $\frac{e^{-x}}{x^{-2}}$ is also legal and also indeterminate, now $\frac{0}{0}$ , but differentiating gives $\frac{-e^{-x}}{-2x^{-3}}$ , which is worse than the start. Choose the arrangement whose derivatives simplify, which here means the polynomial on top.

The general result, and the shape of the graph

The number of passes is the only thing that changes with the power. Each pass drops the exponent by one while the exponential regenerates, so $x^n$ on top needs $n$ passes and finishes as $\frac{n!}{e^x}$ .

\lim_{x \to \infty} x^ne^{-x} = 0 \quad \text{for every } n > 0

The function is not decreasing the whole way, which is worth knowing before trusting a table of values. Its derivative is $\left(2x-x^2\right)e^{-x}$ , zero at $x = 0$ and $x = 2$ .

\frac{d}{dx}\left[x^2e^{-x}\right] = \left(2x-x^2\right)e^{-x}

So the graph climbs to a maximum of $4e^{-2} \approx 0.5413$ at $x = 2$ and decays from there. The limit describes the tail, not the whole story.

The mistakes students make

Answering $\infty$ because $x^2 \to \infty$ , or $0$ because $e^{-x} \to 0$ . Both readings guess at an indeterminate form instead of resolving it.
Applying the rule directly to the product. L'Hopital takes a quotient, so the rewrite is not bookkeeping, it is what makes the rule legal.
Stopping after one pass and reporting $\frac{2x}{e^x}$ , or carrying on past $\frac{2}{e^x}$ where the form is no longer indeterminate.
Differentiating $e^{-x}$ as $e^{-x}$ . The chain rule gives $-e^{-x}$ , and a lost sign there flips the comparison.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

Why is infinity times 0 not simply 0?

Because the factor heading to $0$ may be shrinking more slowly than the other is growing. The product $x^2 \cdot \frac{1}{x}$ has exactly that shape and runs to $\infty$ . Only a comparison of rates settles it, and turning the product into a quotient is what makes the comparison possible.

Which factor should go in the denominator?

The one that gets simpler when it moves. Sending $e^{-x}$ down produces $\frac{x^2}{e^x}$ , where each derivative shrinks the numerator toward a constant. Sending $x^2$ down produces $\frac{e^{-x}}{x^{-2}}$ , where the derivatives grow more complicated. Both are valid; only one finishes.

Does the answer change for $x^{10}e^{-x}$ ?

No, it is still $0$ , though it takes ten passes rather than two and the approach is slower. That function climbs to its maximum at $x = 10$ before turning, so a table sampled below $x = 10$ would suggest the opposite answer.