4.8 L’Hôpital’s Rule

L'Hôpital's Rule and Its Applications

L'Hôpital's Rule gives you a way to evaluate limits that produce indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Instead of struggling with algebraic manipulation, you differentiate the numerator and denominator separately, then re-evaluate the limit. It's one of the most useful techniques in this unit for analyzing function behavior, comparing growth rates, and finding asymptotes.

Applicability of L'Hôpital's Rule

Before applying the rule, you need to confirm the limit actually produces an indeterminate form. Direct substitution that gives you a clean number (like $\frac{3}{5}$) doesn't need L'Hôpital's Rule at all.

The indeterminate forms you'll encounter:

• $\frac{0}{0}$: both numerator and denominator approach 0 (e.g., $\lim_{x \to 0} \frac{\sin x}{x}$)
• $\frac{\infty}{\infty}$: both numerator and denominator approach $\infty$ or $-\infty$ (e.g., $\lim_{x \to \infty} \frac{x^2+1}{x+1}$)
• $0 \cdot \infty$: one factor approaches 0 while the other approaches $\pm\infty$ (e.g., $\lim_{x \to \infty} xe^{-x}$)
• $\infty - \infty$: a difference of two expressions that each approach $\infty$ (e.g., $\lim_{x \to \infty} (\sqrt{x^2+1} - x)$)
• $0^0$, $1^{\infty}$, $\infty^0$: exponential forms where the base and exponent create ambiguity (e.g., $\lim_{x \to 0^+} x^x$)

Only the first two forms ($\frac{0}{0}$ and $\frac{\infty}{\infty}$) can use L'Hôpital's Rule directly. The others need to be rewritten first (more on that below).

Three conditions must hold before you apply the rule:

1. The limit produces an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
2. The numerator and denominator are both differentiable near the point in question
3. The limit of the ratio of derivatives, $\lim \frac{f'(x)}{g'(x)}$, actually exists (or equals $\pm\infty$)

That third condition matters. If the derivative ratio keeps oscillating without settling on a value, L'Hôpital's Rule doesn't apply, even though the original limit might still exist.

Evaluation of Indeterminate Forms

For $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms:

1. Confirm the limit gives an indeterminate form by substituting
2. Differentiate the numerator and denominator separately (this is not the quotient rule)
3. Evaluate the new limit $\lim \frac{f'(x)}{g'(x)}$
4. If the result is still indeterminate, repeat the process

For example, $\lim_{x \to 0} \frac{\sin x}{x}$ gives $\frac{0}{0}$. Differentiating top and bottom: $\lim_{x \to 0} \frac{\cos x}{1} = 1$.

A common mistake: using the quotient rule instead of differentiating numerator and denominator independently. L'Hôpital's Rule says $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$, not $\lim \left(\frac{f}{g}\right)'$.

For $0 \cdot \infty$ forms:

1. Rewrite the product as a quotient by moving one factor to the denominator as its reciprocal. For instance, rewrite $xe^{-x}$ as $\frac{x}{e^x}$
2. Now you have a $\frac{\infty}{\infty}$ form, so apply L'Hôpital's Rule normally

For $\infty - \infty$ forms:

1. Combine the expression into a single fraction (find a common denominator or factor creatively)
2. The resulting fraction should be $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and you can apply L'Hôpital's Rule

For $0^0$, $1^{\infty}$, or $\infty^0$ forms:

1. Set $y = f(x)^{g(x)}$ and take the natural log: $\ln y = g(x) \cdot \ln f(x)$
2. This converts the problem to a $0 \cdot \infty$ form (or similar), which you handle with the techniques above
3. After finding $\lim \ln y = L$, the original limit is $e^L$

Don't forget that last step. A very common error is solving for the limit of $\ln y$ and reporting that as your final answer.

Function Growth Rate Comparisons

L'Hôpital's Rule is especially handy for comparing how fast two functions grow as $x \to \infty$. To compare $f(x)$ and $g(x)$, evaluate:

$\lim_{x \to \infty} \frac{f(x)}{g(x)}$

How to interpret the result:

• Limit = 0: $f(x)$ grows slower than $g(x)$
• Limit = non-zero constant: $f(x)$ and $g(x)$ grow at the same rate
• Limit = $\pm\infty$: $f(x)$ grows faster than $g(x)$

The standard growth rate hierarchy (from slowest to fastest as $x \to \infty$) is:

$\ln x \ll x^n \ll e^x \ll x^x$

For example, $\lim_{x \to \infty} \frac{x^3}{e^x}$. This is $\frac{\infty}{\infty}$, so apply L'Hôpital's Rule three times. Each time the polynomial degree drops by 1, while $e^x$ stays $e^x$. The limit equals 0, confirming that any polynomial grows slower than $e^x$.

This kind of comparison shows up when determining dominant terms in expressions (e.g., for large $x$, $x^3 + 2x^2 + 3x + 4$ behaves like $x^3$) and when analyzing horizontal asymptotes.

Historical Context and Applications

Guillaume de l'Hôpital published this rule in 1696 in the first calculus textbook. The result was actually discovered by Johann Bernoulli, who shared it with l'Hôpital as part of a paid arrangement. The rule applies throughout physics, engineering, and economics wherever you need to evaluate limits involving rates of change or compare long-term behavior of competing models.