💡 Key Idea: When direct substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, take the derivative of the top and bottom separately, then try again!

The Rule (Formal Statement)

Suppose $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (or both approach $\pm\infty$).

If $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

Important: This works for $x \to a$, $x \to a^+$, $x \to a^-$, $x \to \infty$, or $x \to -\infty$!

When to Use L'Hôpital's Rule

Indeterminate Forms That Work

L'Hôpital's Rule applies ONLY to these forms:

1. $\frac{0}{0}$ ← most common
2. $\frac{\infty}{\infty}$ ← also very common

Forms That DON'T Work Directly

These are NOT indeterminate forms for L'Hôpital's Rule:

• $\frac{1}{0}$ (this is $\pm\infty$, not indeterminate)
• $\frac{0}{\infty}$ (this equals 0)
• $\frac{\infty}{0}$ (this is $\pm\infty$)

But wait! We can sometimes convert other indeterminate forms:

• $0 \cdot \infty$ → rewrite as $\frac{0}{0}$ or $\frac{\infty}{\infty}$
• $\infty - \infty$ → combine and rewrite as a fraction
• $0^0$, $1^\infty$, $\infty^0$ → use logarithms first!

How to Apply L'Hôpital's Rule

Step-by-Step Process

Step 1: Try direct substitution first

Step 2: Check if you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$

• If YES → L'Hôpital's Rule applies!
• If NO → use another method

Step 3: Take derivatives of numerator and denominator separately $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

Step 4: Evaluate the new limit

• If you get a number, you're done!
• If you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ again, apply L'Hôpital's Rule again!
• If you get something else, stop and use a different method

Example 1: Basic Application

Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$

Step 1: Direct substitution

$\frac{\sin 0}{0} = \frac{0}{0}$ ← indeterminate form!

Step 2: Apply L'Hôpital's Rule

Take derivatives of top and bottom separately:

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{(\sin x)'}{(x)'}$

$= \lim_{x \to 0} \frac{\cos x}{1}$

Step 3: Evaluate

$= \frac{\cos 0}{1} = \frac{1}{1} = 1$

Answer: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Example 2: Apply Twice

Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

First attempt: Direct substitution

$\frac{1 - \cos 0}{0^2} = \frac{1-1}{0} = \frac{0}{0}$ ← use L'Hôpital's!

Apply L'Hôpital's Rule (first time)

$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{(1-\cos x)'}{(x^2)'}$

$= \lim_{x \to 0} \frac{\sin x}{2x}$

Check: Direct substitution again

$\frac{\sin 0}{2(0)} = \frac{0}{0}$ ← still indeterminate!

Apply L'Hôpital's Rule (second time)

$\lim_{x \to 0} \frac{\sin x}{2x} = \lim_{x \to 0} \frac{(\sin x)'}{(2x)'}$

$= \lim_{x \to 0} \frac{\cos x}{2}$

$= \frac{\cos 0}{2} = \frac{1}{2}$

Answer: $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Example 3: Infinity Form

Evaluate $\lim_{x \to \infty} \frac{\ln x}{x}$

Step 1: Check the form

As $x \to \infty$: $\ln x \to \infty$ and $x \to \infty$

Form: $\frac{\infty}{\infty}$ ← L'Hôpital's applies!

Step 2: Apply L'Hôpital's Rule

$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{(\ln x)'}{(x)'}$

$= \lim_{x \to \infty} \frac{1/x}{1}$

$= \lim_{x \to \infty} \frac{1}{x} = 0$

Answer: $\lim_{x \to \infty} \frac{\ln x}{x} = 0$

Interpretation: Logarithms grow slower than linear functions!

Converting Other Indeterminate Forms

Form: $0 \cdot \infty$

Strategy: Rewrite as a fraction

$\lim_{x \to 0^+} x \ln x$

This is $0 \cdot (-\infty)$ form.

Rewrite: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$

Now it's $\frac{-\infty}{\infty}$ form!

Apply L'Hôpital's: $= \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0$

Form: $\infty - \infty$

Strategy: Combine into a single fraction

$\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{\sin x}\right)$

This is $\infty - \infty$ form.

Combine: $\lim_{x \to 0^+} \frac{\sin x - x}{x \sin x}$

Now it's $\frac{0}{0}$ form → use L'Hôpital's!

Form: $1^\infty$, $0^0$, $\infty^0$

Strategy: Take natural log first, use L'Hôpital's, then exponentiate

$\lim_{x \to 0^+} x^x$

This is $0^0$ form.

Let $y = x^x$, then $\ln y = x \ln x$

Find $\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln x = 0$ (from earlier)

Therefore: $\lim_{x \to 0^+} y = e^0 = 1$

Example: Exponential Form

Evaluate $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$

This is $1^\infty$ form.

Step 1: Take natural log

Let $y = \left(1 + \frac{1}{x}\right)^x$

$\ln y = x \ln\left(1 + \frac{1}{x}\right)$

Step 2: Evaluate limit of $\ln y$

$\lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \ln\left(1 + \frac{1}{x}\right)$

This is $\infty \cdot 0$ form.

Rewrite: $\lim_{x \to \infty} \frac{\ln(1 + 1/x)}{1/x}$ (now $\frac{0}{0}$ form)

Step 3: Apply L'Hôpital's

$= \lim_{x \to \infty} \frac{\frac{1}{1+1/x} \cdot (-1/x^2)}{-1/x^2}$

$= \lim_{x \to \infty} \frac{1}{1+1/x} = 1$

Step 4: Exponentiate

$\lim_{x \to \infty} y = e^{\lim \ln y} = e^1 = e$

Answer: $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$

This is actually the definition of $e$! 🎉

⚠️ Common Mistakes

Mistake 1: Using When It Doesn't Apply

WRONG: Using L'Hôpital's for $\lim_{x \to 0} \frac{x}{1} = 0$

This is NOT indeterminate! Just substitute.

Mistake 2: Taking Derivative of the Whole Fraction

WRONG: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]$ (using quotient rule)

RIGHT: $\frac{f'(x)}{g'(x)}$ (derivatives separately!)

L'Hôpital's says take derivatives of top and bottom separately, not the derivative of the quotient!

Mistake 3: Not Checking the Form

Always verify you have $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying!

Mistake 4: Applying Infinitely

If L'Hôpital's keeps giving $\frac{0}{0}$, you might be in a loop. Try a different method!

Mistake 5: Forgetting to Simplify

Sometimes simplifying algebraically is easier than using L'Hôpital's multiple times.

When NOT to Use L'Hôpital's Rule

Alternative 1: Algebra is Easier

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4$

Yes, L'Hôpital's would work, but factoring is simpler!

Alternative 2: Standard Limits

You should know: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Don't need L'Hôpital's every time!

Alternative 3: The Form Doesn't Match

If you get $\frac{5}{0}$, that's $\infty$ (or $-\infty$), not indeterminate!

Quick Reference Chart

Growth Rates Using L'Hôpital's

L'Hôpital's Rule helps us compare growth rates:

Logarithms grow slower than polynomials: $\lim_{x \to \infty} \frac{\ln x}{x^p} = 0 \text{ for any } p > 0$

Polynomials grow slower than exponentials: $\lim_{x \to \infty} \frac{x^n}{e^x} = 0 \text{ for any } n$

Order from slowest to fastest: $\ln x \ll x^p \ll e^x \ll x!$

Historical Note

Named after Guillaume de l'Hôpital (1661-1704), though it was actually discovered by his teacher Johann Bernoulli!

L'Hôpital paid Bernoulli to teach him calculus and share his discoveries. The rule appeared in L'Hôpital's book, the first calculus textbook ever published!

📝 Practice Strategy

1. Always try direct substitution first - might not be indeterminate!
2. Check the form - only $\frac{0}{0}$ and $\frac{\infty}{\infty}$ work directly
3. Convert other forms to one of the two usable forms
4. Take derivatives separately - not the derivative of the quotient!
5. Simplify when possible - sometimes algebra is faster
6. Be ready to apply multiple times - but watch for loops
7. For exponential forms, take ln first, then exponentiate at the end

Form: $\frac{\infty}{\infty}$ ← L'Hôpital's applies!

Step 2: Apply L'Hôpital's Rule (first time)

$\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{(x^2)'}{(e^x)'}$

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

Step 3: Check the form again

As $x \to \infty$: $2x \to \infty$ and $e^x \to \infty$

Still $\frac{\infty}{\infty}$ form!

Step 4: Apply L'Hôpital's Rule (second time)

$\lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{(2x)'}{(e^x)'}$

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

Step 5: Evaluate

$= \frac{2}{\infty} = 0$

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

Interpretation:

Even though $x^2$ grows to infinity, $e^x$ grows so much faster that the ratio goes to 0.

Growth rate conclusion: Exponential functions grow faster than any polynomial!

In general: $\lim_{x \to \infty} \frac{x^n}{e^x} = 0$ for any positive integer $n$.

So $(\sin x)^x$ is the form $0^0$ (indeterminate!)

Can't use L'Hôpital's directly on this form.

Step 2: Use logarithms

Let $y = (\sin x)^x$

Take natural log of both sides: $\ln y = \ln[(\sin x)^x] = x \ln(\sin x)$

Step 3: Find limit of $\ln y$

$\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln(\sin x)$

This is $0 \cdot (-\infty)$ form.

Step 4: Rewrite as a fraction

$\lim_{x \to 0^+} x \ln(\sin x) = \lim_{x \to 0^+} \frac{\ln(\sin x)}{1/x}$

This is $\frac{-\infty}{\infty}$ form → can use L'Hôpital's!

Step 5: Apply L'Hôpital's Rule

Numerator: $(\ln(\sin x))' = \frac{1}{\sin x} \cdot \cos x = \frac{\cos x}{\sin x} = \cot x$

Denominator: $(1/x)' = -1/x^2$

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

$= \lim_{x \to 0^+} \frac{\cos x}{\sin x} \cdot (-x^2)$

$= \lim_{x \to 0^+} \frac{-x^2 \cos x}{\sin x}$

Step 6: Simplify and apply L'Hôpital's again

$= \lim_{x \to 0^+} \frac{-x^2 \cos x}{\sin x} = \lim_{x \to 0^+} \left(-x \cdot \frac{x\cos x}{\sin x}\right)$

Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we have $\lim_{x \to 0} \frac{x}{\sin x} = 1$

$= \lim_{x \to 0^+} (-x) \cdot \cos x \cdot 1 = 0 \cdot 1 = 0$

Step 7: Exponentiate to find $y$

$\lim_{x \to 0^+} \ln y = 0$

Therefore: $\lim_{x \to 0^+} y = e^0 = 1$

Answer: $\lim_{x \to 0^+} (\sin x)^x = 1$