Figure 5.1: Geometric Intuition of L'Hôpital's Rule at Indeterminate Form $\frac{0}{0}$
1. Indeterminate Forms & Classification
Direct substitution into a limit $\lim_{x \to c} \frac{f(x)}{g(x)}$ may yield an expression whose value cannot be determined from the limits of the individual parts alone. In IB Mathematics AA HL, you are expected to identify and resolve all seven classical indeterminate forms:
Primary Quotients (L'Hôpital Direct)
- $\frac{0}{0}$: Vanishing ratio, dictated by initial derivatives $f'(c)/g'(c)$.
- $\frac{\pm\infty}{\pm\infty}$: Competing growth rates of unbounded functions.
Algebraic Conversions (Quotient First)
- $0 \times \infty$: Invert one factor: $f \cdot g = \frac{f}{1/g} \implies \frac{0}{0}$ or $\frac{\infty}{\infty}$.
- $\infty - \infty$: Combine over a common denominator or multiply by conjugate.
Exponential Powers ($y = f^g$)
- $1^\infty$: Base approaches 1 while power explodes (source of $e$).
- $0^0$ and $\infty^0$: Take $\ln y = g(x)\ln f(x)$ to convert to $0 \times \infty$.
Determinate Forms Warning
Never apply L'Hôpital's Rule to determinate expressions! Common traps: $\frac{0}{\infty} = 0$, $\frac{\infty}{0} = \pm\infty$, $\frac{k}{0} = \pm\infty$ (vertical asymptote, non-existent two-sided limit), $\infty + \infty = \infty$, and $0^\infty = 0$. Applying the rule to these yields erroneous results.
2. L'Hôpital's Rule: Theorem & Examiner Criteria (Syllabus 5.12)
Let $f$ and $g$ be differentiable on an open interval containing $c$ (except possibly at $c$ itself), with $g'(x) \neq 0$ near $c$. If:
$$\lim_{x \to c} f(x) = 0 \quad \text{and} \quad \lim_{x \to c} g(x) = 0 \quad \left(\text{Form } \frac{0}{0}\right)$$ $$\text{OR} \quad \lim_{x \to c} |f(x)| = \infty \quad \text{and} \quad \lim_{x \to c} |g(x)| = \infty \quad \left(\text{Form } \frac{\infty}{\infty}\right)$$Then:
$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$provided the derivative limit exists or is $\pm\infty$. This theorem holds equally for one-sided limits ($x \to c^+$, $x \to c^-$) and infinite limits ($x \to \infty$, $x \to -\infty$).
IB Markscheme Requirement (Guaranteed Method Marks)
IB examiners award a standalone method mark [M1] for explicitly verifying and stating the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before differentiating. Omitting this statement or failing to verify prior to differentiation frequently leads to a loss of subsequent accuracy marks [A0].
3. Rigorous Techniques for All Indeterminate Forms
A. Exponential Indeterminate Powers ($1^\infty$, $0^0$, $\infty^0$)
When evaluating $\lim_{x \to c} [f(x)]^{g(x)}$, use logarithmic transformation:
- Define $y = [f(x)]^{g(x)}$.
- Apply natural logarithms: $\ln y = \ln([f(x)]^{g(x)}) = g(x) \ln f(x)$.
- Evaluate $k = \lim_{x \to c} \ln y = \lim_{x \to c} \frac{\ln f(x)}{1/g(x)}$ using L'Hôpital's Rule on form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
- Recover the original limit using continuity of the exponential function: $$\lim_{x \to c} [f(x)]^{g(x)} = \lim_{x \to c} e^{\ln y} = e^k$$
B. Difference Form $\infty - \infty$ with Conjugate Radicals
For limits involving radical differences such as $\lim_{x \to \infty} (\sqrt{x^2 + ax} - x)$, rationalize by multiplying and dividing by the conjugate $\sqrt{x^2 + ax} + x$:
$$\lim_{x \to \infty} \frac{(\sqrt{x^2 + ax} - x)(\sqrt{x^2 + ax} + x)}{\sqrt{x^2 + ax} + x} = \lim_{x \to \infty} \frac{ax}{\sqrt{x^2 + ax} + x} = \frac{a}{2}$$C. Asymptotic Growth Hierarchy at Infinity
For large $x \to \infty$, functions grow in strictly dominant tiers. In Paper 1 (Non-Calculator), knowing this hierarchy provides instant verification:
$$\ln x \ll x^p \ll a^x \ll x! \ll x^x \quad (p > 0, a > 1)$$Consequently, any power dominates any logarithm ($\lim_{x \to \infty} \frac{(\ln x)^m}{x^p} = 0$), and any exponential dominates any polynomial ($\lim_{x \to \infty} \frac{x^p}{e^{kx}} = 0$ for $k > 0$).
D. L'Hôpital vs. Maclaurin Series Expansions
In AA HL Paper 1, evaluating limits of the form $\lim_{x \to 0} \frac{f(x)}{x^n}$ where $n \ge 3$ via L'Hôpital requires 3 or more tedious differentiation steps with product/chain rules. Substituting known Maclaurin expansions is often 5 times faster:
$$\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5), \quad \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^6), \quad e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \mathcal{O}(x^4)$$For example: $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{x - (x - x^3/6)}{x^3} = \frac{1}{6}$ in a single step!
Worked Examples
Worked Example 1: Repeated L'Hôpital with Second Order Vanishing
Evaluate $\lim_{x \to 0} \frac{e^{3x} - 1 - 3x}{x^2}$.
Solution:
1. Test direct substitution at $x = 0$:
$$\text{Numerator: } e^0 - 1 - 0 = 0, \quad \text{Denominator: } 0^2 = 0 \quad \implies \text{Indeterminate form } \frac{0}{0} \quad \text{[M1]}$$2. Differentiate numerator and denominator separately with respect to $x$:
$$\lim_{x \to 0} \frac{\frac{d}{dx}(e^{3x} - 1 - 3x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{3e^{3x} - 3}{2x} \quad \text{[M1]}$$3. Direct substitution yields $\frac{3(1) - 3}{2(0)} = \frac{0}{0}$. Indeterminate form $\frac{0}{0}$ confirmed.
Apply L'Hôpital's Rule a second time:
4. Evaluate the limit directly:
$$= \frac{9e^0}{2} = \frac{9}{2} \quad \text{[A1]}$$Worked Example 2: Product Form Inversion $0 \times \infty$
Evaluate $\lim_{x \to 0^+} x^2 \ln x$.
Solution:
1. As $x \to 0^+$, $x^2 \to 0$ and $\ln x \to -\infty$, giving indeterminate product $0 \times (-\infty)$.
Rewrite as an indeterminate quotient by placing $x^2$ into the denominator as $x^{-2}$:
2. Apply L'Hôpital's Rule by differentiating numerator and denominator:
$$\lim_{x \to 0^+} \frac{1/x}{-2x^{-3}} = \lim_{x \to 0^+} \left(-\frac{1}{2} \cdot \frac{x^3}{x}\right) = \lim_{x \to 0^+} \left(-\frac{1}{2} x^2\right) \quad \text{[M1]}$$3. Substitute $x = 0$:
$$= -\frac{1}{2}(0)^2 = 0 \quad \text{[A1]}$$Worked Example 3: Difference Form $\infty - \infty$ with Common Denominator
Evaluate $\lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{\sin x} \right)$.
Solution:
1. Direct substitution gives $\infty - \infty$. Combine over a common denominator:
$$\lim_{x \to 0} \frac{\sin x - x}{x \sin x} \quad \left(\text{Form } \frac{0}{0}\right) \quad \text{[M1]}$$2. Apply L'Hôpital's Rule (using product rule on the denominator):
$$\lim_{x \to 0} \frac{\cos x - 1}{(1)\sin x + x\cos x} \quad \text{[M1]}$$3. Direct substitution yields $\frac{1 - 1}{0 + 0} = \frac{0}{0}$. Apply L'Hôpital a second time:
$$\lim_{x \to 0} \frac{-\sin x}{\cos x + \cos x - x\sin x} = \lim_{x \to 0} \frac{-\sin x}{2\cos x - x\sin x} \quad \text{[M1]}$$4. Evaluate at $x = 0$:
$$= \frac{-\sin 0}{2\cos 0 - 0\sin 0} = \frac{0}{2(1) - 0} = 0 \quad \text{[A1]}$$Worked Example 4: Indeterminate Exponential Power $1^\infty$
Evaluate $\lim_{x \to 0} (1 + 3x)^{2/x}$.
Solution:
1. As $x \to 0$, $(1 + 3x) \to 1$ and $\frac{2}{x} \to \pm\infty$, which is indeterminate form $1^\infty$.
Set $y = (1 + 3x)^{2/x}$. Take natural logarithm of both sides:
2. Evaluate $\lim_{x \to 0} \ln y$:
$$\lim_{x \to 0} \frac{2\ln(1 + 3x)}{x} \quad \left(\text{Form } \frac{0}{0}\right) \quad \text{[M1]}$$3. Apply L'Hôpital's Rule:
$$\lim_{x \to 0} \frac{2 \cdot \frac{3}{1 + 3x}}{1} = \lim_{x \to 0} \frac{6}{1 + 3x} = \frac{6}{1 + 0} = 6 \quad \text{[A1]}$$4. Since $\lim_{x \to 0} \ln y = 6$, exponentiate to find the original limit:
$$\lim_{x \to 0} y = e^6 \quad \text{[A1]}$$Worked Example 5: Parameter Determination (Classic 6-Mark Exam Question)
Given that $\lim_{x \to 0} \frac{\cos(ax) + bx^2 - 1}{x^4} = -3$, find the exact values of the non-zero real constants $a$ and $b$.
Solution:
1. At $x = 0$, denominator is 0. For a finite limit to exist, the numerator must also be 0:
$\cos(0) + b(0)^2 - 1 = 1 - 1 = 0$ (verified, form $\frac{0}{0}$).
Apply L'Hôpital's Rule:
2. Form is $\frac{0}{0}$. Apply L'Hôpital a second time:
$$\lim_{x \to 0} \frac{-a^2\cos(ax) + 2b}{12x^2} \quad \text{[M1]}$$3. The denominator is 0 at $x = 0$. For the limit to be finite, the numerator must vanish at $x = 0$:
$$-a^2\cos(0) + 2b = 0 \implies -a^2 + 2b = 0 \implies 2b = a^2 \implies b = \frac{a^2}{2} \quad \text{[A1]}$$4. Substitute $2b = a^2$, giving indeterminate form $\frac{0}{0}$. Apply L'Hôpital a third time:
$$\lim_{x \to 0} \frac{a^3\sin(ax)}{24x} \quad \text{[M1]}$$Apply L'Hôpital a fourth time (or use $\lim_{x \to 0}\frac{\sin(ax)}{x} = a$):
$$\lim_{x \to 0} \frac{a^4\cos(ax)}{24} = \frac{a^4(1)}{24} = \frac{a^4}{24} \quad \text{[A1]}$$5. Equate to the given limit value $-3$:
$$\frac{a^4}{24} = -3 \implies a^4 = -72$$Since $a \in \mathbb{R}$, $a^4 \ge 0$. Thus, for the limit to equal $-3$, there are no real solutions for $a$ if defined as $-3$. If the limit were $+3$:
$a^4 = 72 \implies a = \pm\sqrt[4]{72} = \pm 6^{1/2} \cdot 2^{1/4}$, and $b = \frac{a^2}{2} = \frac{\sqrt{72}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$.
Note: For negative limit values, the numerator must have opposite sign (e.g., $1 - bx^2 - \cos(ax)$). This confirms how sensitive parameter questions are to sign analysis! [R1]
Practice Exercises (10 Tiered Questions)
Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.
Evaluate $\lim_{x \to 0} \frac{\sin(5x)}{2x}$ using L'Hôpital's rule.
Show Full Solution
Direct substitution: $\frac{\sin 0}{0} = \frac{0}{0}$ (indeterminate form). [M1]
Apply L'Hôpital: $\lim_{x \to 0} \frac{5\cos(5x)}{2} = \frac{5\cos(0)}{2} = \frac{5(1)}{2} = \frac{5}{2}$. [A1]
Evaluate $\lim_{x \to \infty} \frac{3x^2 + 1}{5x^2 - 4x}$.
Show Full Solution
Form $\frac{\infty}{\infty}$. [M1]
Apply L'Hôpital: $\lim_{x \to \infty} \frac{6x}{10x - 4} = \frac{\infty}{\infty}$.
Apply L'Hôpital again: $\lim_{x \to \infty} \frac{6}{10} = \frac{3}{5}$. [A1]
Evaluate $\lim_{x \to \infty} \frac{e^{2x}}{x}$.
Show Full Solution
Form $\frac{\infty}{\infty}$. [M1]
Apply L'Hôpital: $\lim_{x \to \infty} \frac{2e^{2x}}{1} = +\infty$. [A1]
Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.
Show Full Solution
Form $\frac{1 - 1}{0} = \frac{0}{0}$. [M1]
First application: $\lim_{x \to 0} \frac{\sin x}{2x}$ (still $\frac{0}{0}$).
Second application: $\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$. [A1]
Evaluate $\lim_{x \to 1} \frac{\ln x}{x^2 - 1}$.
Show Full Solution
Form $\frac{\ln 1}{1 - 1} = \frac{0}{0}$. [M1]
Apply L'Hôpital: $\lim_{x \to 1} \frac{1/x}{2x} = \lim_{x \to 1} \frac{1}{2x^2} = \frac{1}{2}$. [A1]
Evaluate $\lim_{x \to 0^+} x \ln(x)$.
Show Full Solution
Form $0 \times (-\infty)$. Rewrite: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$ (form $\frac{-\infty}{\infty}$). [M1]
Apply L'Hôpital: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$. [A1]
Evaluate $\lim_{x \to 0} \frac{\arctan x - x}{x^3}$.
Show Full Solution
Form $\frac{0 - 0}{0} = \frac{0}{0}$. [M1]
First application: $\lim_{x \to 0} \frac{\frac{1}{1+x^2} - 1}{3x^2} = \lim_{x \to 0} \frac{\frac{1 - (1+x^2)}{1+x^2}}{3x^2} = \lim_{x \to 0} \frac{-x^2}{3x^2(1+x^2)}$. [M1]
Cancel $x^2$: $\lim_{x \to 0} \frac{-1}{3(1+x^2)} = -\frac{1}{3}$. [A1]
Given that $\lim_{x \to 0} \frac{\sin(2x) + a x + b x^3}{x^3} = L$ where $L$ is a finite non-zero constant, determine all possible values of $a$, $b$, and $L$.
Show Full Solution
At $x = 0$, denominator is 0. For limit to be finite, numerator must be 0: $\sin 0 + 0 = 0$ (holds).
Apply L'Hôpital: $\lim_{x \to 0} \frac{2\cos(2x) + a + 3bx^2}{3x^2}$. [M1]
At $x = 0$, denominator is 0. So numerator must be 0:
$2\cos(0) + a + 0 = 2 + a = 0 \implies a = -2$. [A1]
Now limit is: $\lim_{x \to 0} \frac{2\cos(2x) - 2 + 3bx^2}{3x^2}$ (form $\frac{0}{0}$).
Apply L'Hôpital again: $\lim_{x \to 0} \frac{-4\sin(2x) + 6bx}{6x} = \lim_{x \to 0} \left(-\frac{4\sin(2x)}{6x} + b\right)$. [M1]
Since $\lim_{x \to 0} \frac{\sin(2x)}{x} = 2$: $-\frac{4(2)}{6} + b = -\frac{4}{3} + b$.
For any finite non-zero $L$: $L = b - \frac{4}{3}$. (If evaluated through third L'Hôpital: $\frac{-8\cos(2x) + 6b}{6} = \frac{-8 + 6b}{6} = b - \frac{4}{3}$). [A1]
All solutions: $a=-2$, $b\in\mathbb{R}\setminus\{\frac43\}$ and $L=b-\frac43$. The non-zero condition excludes $b=\frac43$. There is no unique value of $b$ unless the value of $L$ is supplied.
Evaluate $\lim_{x \to 0} (1 + 3x)^{1/x}$ using natural logarithms.
Show Full Solution
Let $y = (1 + 3x)^{1/x}$. Then $\ln y = \frac{\ln(1 + 3x)}{x}$. [M1]
Evaluate $\lim_{x \to 0} \frac{\ln(1 + 3x)}{x}$ (form $\frac{0}{0}$): [M1]
Apply L'Hôpital: $\lim_{x \to 0} \frac{\frac{3}{1+3x}}{1} = 3$. [A1]
Since $\lim_{x \to 0} \ln y = 3$, $\lim_{x \to 0} y = e^3$. [A1]
Evaluate $\lim_{x \to \infty} \left( \sqrt{x^2 + 4x} - x \right)$.
Show Full Solution
Form $\infty - \infty$. Multiply by conjugate: [M1]
$\lim_{x \to \infty} \frac{(\sqrt{x^2 + 4x} - x)(\sqrt{x^2 + 4x} + x)}{\sqrt{x^2 + 4x} + x} = \lim_{x \to \infty} \frac{(x^2 + 4x) - x^2}{\sqrt{x^2 + 4x} + x} = \lim_{x \to \infty} \frac{4x}{\sqrt{x^2 + 4x} + x}$. [M1]
Divide numerator and denominator by $x$:
$= \lim_{x \to \infty} \frac{4}{\sqrt{1 + 4/x} + 1} = \frac{4}{\sqrt{1 + 0} + 1} = \frac{4}{2} = 2$. [A1]
IB Formula Booklet References (Topic 5)
Syllabus 5.12: L’Hôpital’s Rule
$$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$Condition: $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$, OR $\lim_{x \to c} f(x) = \pm\infty$ and $\lim_{x \to c} g(x) = \pm\infty$.
Key Formulas & Identities
1. Converting Product Form $0 \times \infty$
$$f(x)g(x) = \frac{f(x)}{1/g(x)} \quad \left(\frac{0}{0}\right)$$2. Growth Hierarchy at Infinity
$$\ln x \ll x^k \ll e^x \quad (k > 0)$$Common Exam Pitfalls
Never Apply Quotient Rule to L'Hôpital
L'Hôpital's rule requires differentiating the numerator and denominator separately: $\frac{f'(x)}{g'(x)}$. Do NOT use the quotient rule $\frac{f'g - fg'}{g^2}$!
Check the Condition FIRST
Always state "Form $\frac{0}{0}$" or "Form $\frac{\infty}{\infty}$". Applying L'Hôpital to a determinate limit (e.g. $\frac{1}{0}$ or $\frac{2}{3}$) will yield completely incorrect answers.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
Which of the following is an indeterminate form where L'Hôpital's rule can be directly applied?
What is $\lim_{x \to 0} \frac{e^x - 1}{x}$?
What is $\lim_{x \to \infty} \frac{\ln(x)}{x^2}$?
If direct substitution gives $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{2}{0}$, one should:
Section B: Examination-Style Free Response Questions
Consider the limit $L = \lim_{x \to 0} \frac{\cos(2x) - 1 + 2x^2}{x^4}$.
(a) State the indeterminate form of the limit at $x = 0$. [1 mark]
(b) By repeated application of L’Hôpital’s rule, find the value of $L$. [5 marks]
Show Full Markscheme
(a) Indeterminate form: [1 mark]
At $x = 0$: $\cos(0) - 1 + 0 = 0$, denominator $0^4 = 0$. Form is $\frac{0}{0}$ [A1].
(b) Repeated L'Hôpital: [5 marks]
1st: $\lim_{x \to 0} \frac{-2\sin(2x) + 4x}{4x^3}$ (form $\frac{0}{0}$) [M1].
2nd: $\lim_{x \to 0} \frac{-4\cos(2x) + 4}{12x^2}$ (form $\frac{-4+4}{0} = \frac{0}{0}$) [M1].
3rd: $\lim_{x \to 0} \frac{8\sin(2x)}{24x}$ (form $\frac{0}{0}$) [M1].
4th: $\lim_{x \to 0} \frac{16\cos(2x)}{24} = \frac{16(1)}{24} = \frac{2}{3}$ [A1].
$L = \frac{2}{3}$ [A1].
Let $f(x) = \frac{x - \sin x}{x^3}$ for $x \neq 0$.
(a) Find $\lim_{x \to 0} f(x)$. [3 marks]
(b) Hence define $f(0)$ so that $f$ is continuous at $x = 0$. [1 mark]
(c) Find $\lim_{x \to \infty} f(x)$. [3 marks]
Show Full Markscheme
(a) Limit at 0: [3 marks]
Form $\frac{0 - 0}{0} = \frac{0}{0}$ [M1].
1st: $\lim_{x \to 0} \frac{1 - \cos x}{3x^2}$ (form $\frac{0}{0}$) [M1].
2nd: $\lim_{x \to 0} \frac{\sin x}{6x} = \frac{1}{6}$ [A1].
(b) Value for continuity: [1 mark]
$f(0) = \lim_{x \to 0} f(x) = \frac{1}{6}$ [A1].
(c) Limit at infinity: [3 marks]
$f(x) = \frac{x}{x^3} - \frac{\sin x}{x^3} = \frac{1}{x^2} - \frac{\sin x}{x^3}$ [M1].
As $x \to \infty$, $\frac{1}{x^2} \to 0$.
Since $-1 \le \sin x \le 1$, $-\frac{1}{x^3} \le \frac{\sin x}{x^3} \le \frac{1}{x^3}$.
By the squeeze theorem, $\lim_{x \to \infty} \frac{\sin x}{x^3} = 0$ [M1].
Therefore $\lim_{x \to \infty} f(x) = 0 - 0 = 0$ [A1].