L’Hospital’s Rule and Indeterminate Forms for MATH 139
Exam Relevance for MATH 139
Coverage varies by instructor. If covered, use for 0/0 or infinity/infinity forms only.
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What is L'Hôpital's Rule?
L'Hôpital was a French mathematician (pronounced "loh-pee-TAHL") who published this powerful technique for evaluating tricky limits.
L'Hôpital's Rule is a shortcut for evaluating limits that give you $\frac{0}{0}$ or $\frac{\infty}{\infty}$ when you try direct substitution.
The Big Idea: If plugging in gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, take the derivative of the top and bottom separately, then try again.
⚠️ A word of caution: Some courses limit your use of L'Hôpital's Rule, and exam questions may specifically ask you to evaluate limits without using it. Make sure you also know algebraic techniques (factoring, rationalizing, trig identities) — don't depend on L'Hôpital for every limit!
What Are Indeterminate Forms?
An indeterminate form is an expression that doesn't tell you what the limit actually is — it could be anything!
The Seven Indeterminate Forms
| Form | What it looks like |
|---|---|
| $\frac{0}{0}$ | Both numerator and denominator → 0 |
| $\frac{\infty}{\infty}$ | Both → ∞ (or both → −∞) |
| $0 \cdot \infty$ | One factor → 0, other → ∞ |
| $\infty - \infty$ | Subtracting two things that both → ∞ |
| $0^0$ | Base → 0, exponent → 0 |
| $1^\infty$ | Base → 1, exponent → ∞ |
| $\infty^0$ | Base → ∞, exponent → 0 |
L'Hôpital's Rule directly handles: $\frac{0}{0}$ and $\frac{\infty}{\infty}$
The others require conversion to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first (we'll show how).
L'Hôpital's Rule — The Formula
If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (or both → ∞), then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$
provided the limit on the right exists (or is ±∞).
Critical Rules
- Check first! Only use L'Hôpital when you have $\frac{0}{0}$ or $\frac{\infty}{\infty}$
- Differentiate separately — NOT the quotient rule! Take $f'(x)$ and $g'(x)$ individually
- You may need to apply it multiple times if you still get an indeterminate form
- Works for one-sided limits ($x \to a^+$ or $x \to a^-$) and limits at infinity ($x \to \infty$)
Evaluate $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}$
Step 1: Check the form
Direct substitution: $\frac{\sin(0)}{0} = \frac{0}{0}$ ✓ Indeterminate!
Step 2: Apply L'Hôpital's Rule
$$\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = \frac{1}{1}$$
$$\boxed{1}$$
Evaluate $\displaystyle \lim_{x \to \infty} \frac{x^2}{e^x}$
Step 1: Check the form
As $x \to \infty$: numerator → ∞, denominator → ∞. Form: $\frac{\infty}{\infty}$ ✓
Step 2: Apply L'Hôpital's Rule
$$\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}$$
Still $\frac{\infty}{\infty}$! Apply again:
$$= \lim_{x \to \infty} \frac{2}{e^x} = \frac{2}{\infty} = 0$$
$$\boxed{0}$$
Insight: Exponentials grow faster than any polynomial!
Evaluate $\displaystyle \lim_{x \to 0} \frac{\arctan(x)}{x}$
Step 1: Check the form
$\frac{\arctan(0)}{0} = \frac{0}{0}$ ✓
Step 2: Apply L'Hôpital's Rule
Recall: $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$
$$\lim_{x \to 0} \frac{\arctan(x)}{x} = \lim_{x \to 0} \frac{\frac{1}{1+x^2}}{1} = \frac{1}{1+0} = 1$$
$$\boxed{1}$$
Evaluate $\displaystyle \lim_{x \to 0} \frac{\sinh(x) - x}{x^3}$
Step 1: Check the form
$\sinh(0) = 0$, so numerator = $0 - 0 = 0$, denominator = $0$. Form: $\frac{0}{0}$ ✓
Step 2: Apply L'Hôpital's Rule
$$\lim_{x \to 0} \frac{\sinh(x) - x}{x^3} = \lim_{x \to 0} \frac{\cosh(x) - 1}{3x^2}$$
Check: $\frac{\cosh(0) - 1}{0} = \frac{1-1}{0} = \frac{0}{0}$ — apply again!
$$= \lim_{x \to 0} \frac{\sinh(x)}{6x}$$
Still $\frac{0}{0}$! One more time:
$$= \lim_{x \to 0} \frac{\cosh(x)}{6} = \frac{\cosh(0)}{6} = \frac{1}{6}$$
$$\boxed{\frac{1}{6}}$$
Evaluate $\displaystyle \lim_{x \to 0^+} \frac{\ln(x)}{1/x}$
Step 1: Check the form
As $x \to 0^+$: $\ln(x) \to -\infty$ and $\frac{1}{x} \to +\infty$
Form: $\frac{-\infty}{\infty}$ ✓ (L'Hôpital works!)
Step 2: Apply L'Hôpital's Rule
$$\lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{1/x \cdot x^2}{-1} = \lim_{x \to 0^+} \frac{x}{-1} = \frac{0}{-1}$$
$$\boxed{0}$$
Converting Other Indeterminate Forms
Type: $0 \cdot \infty$
Strategy: Rewrite as a fraction to get $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Evaluate $\displaystyle \lim_{x \to 0^+} x \ln(x)$
Step 1: Identify the form
As $x \to 0^+$: $x \to 0$ and $\ln(x) \to -\infty$
Form: $0 \cdot (-\infty)$ — indeterminate but can't use L'Hôpital directly!
Step 2: Rewrite as a fraction
$$x \ln(x) = \frac{\ln(x)}{1/x}$$
Now as $x \to 0^+$: $\frac{-\infty}{\infty}$ ✓
Step 3: Apply L'Hôpital's Rule
$$\lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$
$$\boxed{0}$$
Type: $1^\infty$, $0^0$, $\infty^0$ (Exponential Indeterminate Forms)
Strategy: Use $y = f(x)^{g(x)}$, take $\ln$ of both sides, evaluate the limit, then exponentiate.
Evaluate $\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$
Step 1: Identify the form
As $x \to \infty$: base → $1 + 0 = 1$, exponent → $\infty$
Form: $1^\infty$ — indeterminate!
Step 2: Take the natural log
Let $y = \left(1 + \frac{1}{x}\right)^x$
$$\ln(y) = x \ln\left(1 + \frac{1}{x}\right)$$
Step 3: Evaluate $\lim_{x \to \infty} \ln(y)$
This is $\infty \cdot 0$ form. Rewrite:
$$\ln(y) = \frac{\ln\left(1 + \frac{1}{x}\right)}{1/x}$$
As $x \to \infty$: $\frac{0}{0}$ ✓
Step 4: Apply L'Hôpital's Rule
$$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{1/x} = \lim_{x \to \infty} \frac{\frac{1}{1+1/x} \cdot (-1/x^2)}{-1/x^2}$$
$$= \lim_{x \to \infty} \frac{1}{1 + 1/x} = \frac{1}{1+0} = 1$$
Step 5: Exponentiate
$\ln(y) \to 1$, so $y \to e^1$
$$\boxed{e}$$
Evaluate $\displaystyle \lim_{x \to \infty} x^{1/x}$
Step 1: Identify the form
As $x \to \infty$: base → $\infty$, exponent → $0$
Form: $\infty^0$ — indeterminate!
Step 2: Take the natural log
Let $y = x^{1/x}$
$$\ln(y) = \frac{1}{x} \ln(x) = \frac{\ln(x)}{x}$$
Step 3: Evaluate
As $x \to \infty$: $\frac{\infty}{\infty}$ ✓
$$\lim_{x \to \infty} \frac{\ln(x)}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$$
Step 4: Exponentiate
$\ln(y) \to 0$, so $y \to e^0 = 1$
$$\boxed{1}$$
Tips for L'Hôpital's Rule
- Always verify the indeterminate form before applying
- Simplify first if possible — sometimes algebra avoids the need for L'Hôpital
- Don't use the quotient rule — differentiate top and bottom separately
- If stuck in a loop, try a different approach (algebra, trig identity, etc.)
- For exponential forms, remember: take ln → evaluate → exponentiate back
Common Mistakes and Misunderstandings
❌ Mistake: Using L'Hôpital when you don't have an indeterminate form
Wrong: $\displaystyle \lim_{x \to 2} \frac{x^2}{x+1} = \lim_{x \to 2} \frac{2x}{1} = 4$
Why it's wrong: Direct substitution gives $\frac{4}{3}$, which is NOT indeterminate. L'Hôpital doesn't apply!
Correct: $\displaystyle \lim_{x \to 2} \frac{x^2}{x+1} = \frac{4}{3}$
❌ Mistake: Using the quotient rule instead of differentiating separately
Wrong: $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$
Why it's wrong: L'Hôpital says to take $f'(x)$ and $g'(x)$ separately, NOT the derivative of the whole fraction.
Correct: $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$
❌ Mistake: Forgetting to check if the new limit exists
Wrong: Applying L'Hôpital repeatedly without checking if each step is valid.
Why it's wrong: L'Hôpital only works if the resulting limit exists (or is ±∞).
Correct: After each application, verify you still have an indeterminate form OR that the limit can be evaluated.
L'Hôpital's Rule
When direct substitution gives 0/0 or ∞/∞, take the derivative of the numerator and denominator separately (NOT the quotient rule!), then evaluate the limit again.
Variables:
- $f(x)$:
- The numerator function
- $g(x)$:
- The denominator function
- $f'(x)$:
- Derivative of the numerator
- $g'(x)$:
- Derivative of the denominator
- $a$:
- The point where we're taking the limit (can also be ±∞)
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