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L’Hospital’s Rule

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

This reading includes verified answer corrections and explicitly stated clarifications.

Revisiting Limits

No, we are not done with limits yet. They are so intertwined with derivatives that we can’t possibly imagine learning more about derivatives without also expanding our knowledge of limits! For example, how would we solve $lim_{x \to 0} - 24 x \cdot \ln x$ ? One of the terms approaches 0, the other approaches -∞! 0 times anything is 0, but ∞ times anything is ∞! What happens?!? Actually, zero wins. Infinity gets KOed. You might ask, why? How? This is because of a famous mathematician called L’Hospital (low-pee-TAAL). His rule allows us to determine the limit of indeterminate forms, such as:

$0 \cdot \infty$

$- \frac{\infty}{\infty}$

$\frac{\infty}{\infty}$

$\frac{0}{0}$

$0 \cdot - \infty$

$0^{0}$

$1^{\infty}$

L’Hospital’s Rule applies directly to a quotient with limiting form 0/0 or ∞/∞. Suppose f and g are differentiable on a punctured interval on the side of approach, g′ is nonzero there, and the derivative quotient f′/g′ has a limit (finite or infinite). Under those hypotheses, the original quotient has the same limit. For a two-sided limit, check both sides. Other indeterminate forms must first be transformed into an eligible quotient.

$lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ This equality is conditional on the hypotheses just stated; it is not the quotient rule for derivatives.

We may apply L’Hospital’s Rule again only if the new quotient still has form 0/0 or ∞/∞ and the hypotheses hold again. If the derivative quotient has no limit, the rule gives no conclusion by itself. Check each step instead of differentiating indefinitely. Let’s try it on an eligible example.

Example Problem 5.3.1

Find $lim_{x \to \infty} \frac{x^{5}}{e^{x}}$

The indeterminate form we encounter if we try to take this limit is ∞/∞. So let’s use L’Hospital’s Rule. We get
$lim_{x \to \infty} \frac{\frac{d}{d x} x^{5}}{\frac{d}{d x} e^{x}} = lim_{x \to \infty} \frac{5 x^{4}}{e^{x}}$
However, this is another indeterminate form (∞/∞), so let’s take the derivative again. This gives us:

$lim_{x \to \infty} \frac{20 x^{3}}{e^{x}}$ You can see that we will keep getting indeterminate forms and keep using L’Hospital’s Rule until the power of x reduces to 0. This gives us the result

$lim_{x \to \infty} \frac{5!}{e^{x}} = lim_{x \to \infty} \frac{120}{e^{x}} = 0$

The limit here is 0 because as x approaches ∞, the denominator approaches ∞, and any constant divided by ∞ is 0.

For a nonnegative integer n, the quotient xⁿ/eˣ has form ∞/∞ as x → ∞ when n > 0. Repeated applications of L’Hospital’s Rule, with its hypotheses checked, reduce the numerator to the constant n! while the exponential denominator remains eˣ. The resulting limit is 0. This finite differentiation argument proves the result for nonnegative integer powers, not by itself for every real exponent.

$lim_{x \to \infty} \frac{x^{n}}{e^{x}} = 0$ for each fixed real n. To extend the integer result, choose a nonnegative integer m greater than n. For x ≥ 1, 0 < xⁿ/eˣ ≤ xᵐ/eˣ. The upper bound tends to 0 by the integer argument, so the squeeze theorem proves the real-power result.

Let’s try another interesting problem that was mentioned earlier.

Example Problem 5.3.2

Find $lim_{x \to 0^{+}} x^{x}$ Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

The trickiest part here is rewriting the limit as
$lim_{x \to 0} e^{\ln x^{x}} = lim_{x \to 0} e^{x \cdot \ln x}$

Now we’re stuck. We don’t know what the exponent equals as it approaches 0, so let’s prove its limit separately to figure it out.

$lim_{x \to 0} x \ln x = lim_{x \to 0} \frac{\ln x}{\frac{1}{x}}$ (we can rewrite it this way for the same reason that 2*4=2/(1/4), just replace 4 with x, and 2 with ln(x). Now we can take L’Hospital’s Rule for the above limit.

$lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{x}} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} \ln x}{\frac{d}{d x} (\frac{1}{x})} = lim_{x \to 0^{+}} \frac{\frac{1}{x}}{- \frac{1}{x^{2}}} = lim_{x \to 0^{+}} (- x) = 0$

$lim_{x \to 0} x \cdot \ln x = 0$

Let’s plug this back into the original limit.

We already found that x ln(x) → 0 as x → 0⁺. Now use continuity of the exponential function: e^{x ln(x)} → e⁰ = 1. L’Hospital’s Rule was used to evaluate the exponent’s limit; this final step uses continuity.

$lim_{x \to 0^{+}} e^{x \cdot \ln x} = e^{{lim}_{x \to 0^{+}} x \cdot \ln x} = e^{0} = 1$
Therefore, Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

$lim_{x \to 0^{+}} x^{x} = 1$ Use x > 0 for the real logarithmic definition xˣ = e^{x ln(x)}.

Glossary and deeper reading · 5 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Limit
A limit describes the value approached by outputs as inputs approach a specified target or grow without bound. For an input approaching a finite target, its value at that exact input does not determine the limit. For example, (x²−1)/(x−1) approaches 2 as x approaches 1 even though the original expression is undefined at 1. A finite two-sided limit requires the left and right approaches to agree.
Read more: 1.2 · Limit →
Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →
L’Hospital’s rule
L’Hospital’s rule can evaluate certain quotient limits with the indeterminate forms 0/0 or infinity divided by infinity. Its hypotheses require appropriate differentiability near the target, a nonzero denominator derivative there, and a suitable limit of the derivative quotient. For example, lim(x→0) sin x/x equals lim(x→0) cos x/1=1. Differentiating numerator and denominator separately here is a limit theorem, not the quotient rule for derivatives.
Read more: 4.7 · L’Hospital’s rule →
Differentiability
Differentiability means that a function has a finite derivative at the point under discussion. Differentiability at a point implies continuity there, but continuity alone does not imply differentiability. For example, |x| is continuous at 0 but its left and right tangent slopes differ, so it is not differentiable there. Corners, cusps, vertical tangents, and discontinuities can prevent a finite derivative.
Read more: 2.4 · Differentiability →
Quotient rule
The quotient rule differentiates a quotient where the denominator is nonzero and both functions are differentiable. It gives (f/g)′=(f′g−fg′)/g². For example, differentiating x/(x+1) produces 1/(x+1)² for x≠−1. The order of the subtraction and the denominator square both matter.
Read more: 2.9 · Quotient rule →

Pop Quiz 5.3.1 A visit to L’Hospital

Limit

Derivative

L’Hospital’s rule

Differentiability

Quotient rule

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