AP Calculus AB and BC

Limit of x/e^x as x Approaches Infinity Is 0

Q: Does this mean y = 0 is a horizontal asymptote?

On the right only. The curve flattens onto the x-axis from above as x \to \infty, so y = 0 is a horizontal asymptote there. Heading left the function dives to -\infty, so that side has no asymptote at all.

Q: Can this be done without L'Hopital's rule?

Yes, with a squeeze. For x > 0 every term of the series for e^x is positive, so e^x > \frac{x^2}{2} and therefore 0 < \frac{x}{e^x} < \frac{2}{x}. Both bounds go to 0, so the function between them does too.

Q: What about \frac{x^{100}}{e^x}?

Still 0. One hundred passes reduce the numerator to the constant 100! over e^x, which collapses. The exponential beats every fixed power eventually, however large the power and however late the crossover: \frac{x^{100}}{e^x} keeps climbing until x = 100 before it turns.

The limit of x over e to the x as x approaches infinity is 0. Substitution gives infinity over infinity, an indeterminate form, so L'Hopital's rule applies: differentiating top and bottom separately leaves 1 over e to the x, which goes to 0. The exponential outgrows the linear numerator.

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

Settled by L'Hopital's rule.

Applying the rule once

Check the hypotheses before differentiating anything. Both $x$ and $e^x$ are unbounded as $x \to \infty$ , both are differentiable, and $\frac{d}{dx}\left[e^x\right] = e^x$ is never $0$ . The $\frac{\infty}{\infty}$ form is one of the two the rule accepts.

\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{\frac{d}{dx}[x]}{\frac{d}{dx}\left[e^x\right]} = \lim_{x \to \infty} \frac{1}{e^x}

The new quotient is no longer indeterminate. A fixed numerator over an unbounded denominator collapses, so one pass was enough, and that is the signal to stop.

\lim_{x \to \infty} \frac{1}{e^x} = 0

$x$	$e^x$	$\frac{x}{e^x}$
$5$	$148.4$	$0.0337$
$10$	$22{,}026$	$0.000454$
$20$	$485{,}165{,}195$	$0.0000000412$

By $x = 20$ the numerator has reached $20$ while the denominator has passed $485$ million. The race is not close, and it only gets worse from there.

Why substitution fails

Pushing $\infty$ through the fraction gives $\frac{\infty}{\infty}$ , a form that records no information about the outcome. Both parts grow without bound, and the answer turns entirely on which grows faster, which is exactly what the form suppresses.

The tempting reading is that the two infinities cancel to $1$ . They do not. The quotient $\frac{x^2}{x}$ has the same form and runs to $\infty$ , while $\frac{x}{x^2}$ has it too and goes to $0$ .

Unbounded is not the same as fast

At $x = 20$ the numerator has grown by a factor of $20$ from its value at $1$ , while $e^x$ has grown by a factor of about $178$ million. Both head to $\infty$ , and the indeterminate form is precisely the statement that this comparison has not been made yet.

The growth hierarchy this fixes

Nothing here was special to the first power. Each pass of the rule drops the exponent by one while the denominator regenerates itself, so a numerator of $x^n$ needs $n$ passes and ends as the constant $n!$ over $e^x$ .

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

Pairing that with the logarithm result gives the ordering that settles most end-behaviour questions on sight.

\ln x \ll x^n \ll e^x \quad \text{as } x \to \infty

Flipping the fraction flips the answer, since a positive function tending to $0$ has an unbounded reciprocal.

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

The mistakes students make

Differentiating with the quotient rule instead of taking the two pieces separately. L'Hopital replaces $\frac{f}{g}$ by $\frac{f'}{g'}$ , never by $\left(\frac{f}{g}\right)'$ .
Applying the rule a second time out of habit. After one pass the expression is $\frac{1}{e^x}$ , which is not indeterminate, so the rule no longer applies.
Reading $\frac{\infty}{\infty}$ as $1$ . The form is indeterminate, and the two comparisons above show it can produce anything.
Assuming the other end behaves the same way. As $x \to -\infty$ the quotient is $xe^{-x}$ with $x$ negative and $e^{-x}$ unbounded, so it runs to $-\infty$ and nothing is indeterminate about it.

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

Does this mean $y = 0$ is a horizontal asymptote?

On the right only. The curve flattens onto the $x$ -axis from above as $x \to \infty$ , so $y = 0$ is a horizontal asymptote there. Heading left the function dives to $-\infty$ , so that side has no asymptote at all.

Can this be done without L'Hopital's rule?

Yes, with a squeeze. For $x > 0$ every term of the series for $e^x$ is positive, so $e^x > \frac{x^2}{2}$ and therefore $0 < \frac{x}{e^x} < \frac{2}{x}$ . Both bounds go to $0$ , so the function between them does too.

What about $\frac{x^{100}}{e^x}$ ?

Still $0$ . One hundred passes reduce the numerator to the constant $100!$ over $e^x$ , which collapses. The exponential beats every fixed power eventually, however large the power and however late the crossover: $\frac{x^{100}}{e^x}$ keeps climbing until $x = 100$ before it turns.