IB Mathematics AA AHL The evaluation of limits Study Notes - New Syllabus
IB Mathematics AA AHL Informal ideas of limit, continuity and convergence Study Notes
LEARNING OBJECTIVE
- The evaluation of limits
Key Concepts:
- Evaluation of limits
- L’Hôpital’s rule.
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
Evaluating Limits Using L’Hôpital’s Rule and Maclaurin Series
Evaluating Limits Using L’Hôpital’s Rule and Maclaurin Series
L’Hôpital’s Rule is used to evaluate limits that lead to indeterminate forms like 0/0 or ∞/∞.
Rule Statement:
If limx → a (f(x)/g(x)) gives an indeterminate form, and f and g are differentiable near a with g'(x) ≠ 0, then:
$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $
Repeat if needed until the limit resolves to a determinate form.
Indeterminate Forms: \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \)
Indeterminate forms arise in limit problems where direct substitution leads to an ambiguous expression. The two most common forms are:
- \( \frac{0}{0} \): Occurs when both numerator and denominator approach zero. It requires further simplification, often by factoring, rationalizing, using identities, or applying L’Hôpital’s Rule.
- \( \frac{\infty}{\infty} \): Occurs when both numerator and denominator grow without bound. Use simplification, dominant term comparison, or L’Hôpital’s Rule to evaluate the limit.
These forms do not have a definite value unless further analysis is done.
Maclaurin Series Expansion
Maclaurin series can simplify limit evaluation by expanding functions about \( x = 0 \). Useful expansions:
- \( \sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \cdots \)
- \( \cos x = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \cdots \)
- \( e^x = 1 + x + \frac{x^2}{2!} + \cdots \)
- \( \ln(1 + x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \cdots \)
Use series expansions to cancel or approximate dominant terms and evaluate the limit.
Example
Evaluate \( \lim_{x \to 0} \frac{\sin x}{x} \)
▶️ Answer/Explanation
This is an indeterminate form \( \frac{0}{0} \), so apply L’Hôpital’s Rule:
Differentiate numerator and denominator:
Example
Evaluate \( \lim_{x \to 0} \frac{1 – \cos x}{x^2} \)
▶️ Answer/Explanation
$ \cos x = 1 – \frac{x^2}{2!} + \cdots $
Example
Evaluate \( \lim_{x \to 1} \frac{x^2 – 1}{x – 1} \)
▶️ Answer/Explanation
Substituting \( x = 1 \) gives \( \frac{0}{0} \), an indeterminate form.
Factor numerator:
Example
Evaluate \( \lim_{x \to \infty} \frac{3x^2 + 5}{2x^2 – 7x} \)
▶️ Answer/Explanation
Substituting \( x \to \infty \) gives \( \frac{\infty}{\infty} \), an indeterminate form.
Divide numerator and denominator by \( x^2 \) (the highest power):
Horizontal Asymptotes
Horizontal Asymptotes
A horizontal asymptote describes the behavior of a function \( f(x) \) as \( x \to \infty \) or \( x \to -\infty \). It represents a horizontal line \( y = L \) that the function approaches but does not necessarily touch or cross.
Key Concept:
If \( \lim_{x \to \infty} f(x) = L \) or \( \lim_{x \to -\infty} f(x) = L \), then the line \( y = L \) is a horizontal asymptote.
Rational Functions:
Let \( f(x) = \dfrac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. The horizontal asymptotes depend on the degrees of \( P(x) \) and \( Q(x) \):
- If degree of \( P(x) < \) degree of \( Q(x) \), then \( y = 0 \) is the horizontal asymptote.
- If degree of \( P(x) = \) degree of \( Q(x) \), then \( y = \frac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} \).
- If degree of \( P(x) > \) degree of \( Q(x) \), then there is no horizontal asymptote (there may be an oblique/slant asymptote instead).
Methods:
- Use algebraic simplification (e.g., divide numerator and denominator by the highest power of \( x \)).
- Apply limits: \( \lim_{x \to \pm\infty} f(x) \).
- Use L’Hôpital’s Rule if indeterminate forms like \( \frac{\infty}{\infty} \) occur.
Example
Find the horizontal asymptote of \( f(x) = \dfrac{3x^2 + 5x – 1}{x^2 – 2} \)
▶️ Answer/Explanation
Since the degrees of the numerator and denominator are both 2, use the leading coefficients:
$ \lim_{x \to \infty} \frac{3x^2 + 5x – 1}{x^2 – 2} = \frac{3}{1} = 3 $
Repeated Use of L’Hôpital’s Rule
Repeated Use of L’Hôpital’s Rule
L’Hôpital’s Rule can be applied repeatedly if the limit remains in an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) after the first application.
Procedure:
- Confirm that the limit yields an indeterminate form.
- Apply L’Hôpital’s Rule: Differentiate the numerator and denominator separately.
- Re-check the limit. If it’s still indeterminate, apply the rule again.
- Repeat until a determinate form is achieved.
NOTE:
L’Hôpital’s Rule only applies when the original limit is truly in an indeterminate form and the derivatives involved exist near the point of interest.
Evaluate \( \lim_{x \to 0} \frac{1 – \cos x}{x^2} \)
▶️ Answer/Explanation
First derivative: $ \frac{d}{dx}[1 – \cos x] = \sin x,\quad \frac{d}{dx}[x^2] = 2x $
Second derivative: $ \frac{d}{dx}[\sin x] = \cos x,\quad \frac{d}{dx}[2x] = 2 $
