IB Mathematics AA Concept of a limit Study Notes
IB Mathematics AA Concept of a limit Study Notes
IB Mathematics AA Concept of a limit Notes Offer a clear explanation of Use of Concept of a limit, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Concept of a limit.
Introduction to the Concept of a Limit
Introduction to the Concept of a Limit
A limit describes the value that a function \( f(x) \) approaches as the input \( x \) gets closer to a particular point.
\( \lim_{x \to a} f(x) = L \)
This means that as \( x \) gets arbitrarily close to \( a \), \( f(x) \) gets arbitrarily close to \( L \).
Important points:
- The limit depends on the behavior of \( f(x) \) near \( x = a \), not necessarily at \( x = a \).
- The limit may exist even if \( f(a) \) is undefined.
- Limits can be found by direct substitution, simplification, or graphing.
Notation: Left-hand limit: \( \lim_{x \to a^-} f(x) \) Right-hand limit: \( \lim_{x \to a^+} f(x) \) If both are equal, the limit exists at \( a \).
Example:
Find the limit:
\( \lim_{x \to 2} \frac{x^2 – 4}{x – 2} \)
▶️ Answer/Explanation
Factor the numerator
\( \frac{(x – 2)(x + 2)}{x – 2} \)
Simplify
\( = x + 2 \), for \( x \neq 2 \)
Substitute \( x = 2 \)
\( = 2 + 2 = 4 \)
\( \lim_{x \to 2} \frac{x^2 – 4}{x – 2} = 4 \)
Estimating the Value of a Limit from a Table and Graph
Estimating the Value of a Limit from a Table and Graph
When it is difficult or impossible to compute a limit algebraically, we can estimate it:
Using a table:
Calculate \( f(x) \) for values of \( x \) that approach \( a \) from both sides. If \( f(x) \) approaches the same number, that is the estimated limit.
Notation: If the table suggest the same value, we write:
\( \lim_{x \to a} f(x) = L \)
Example:
Estimate \( \lim_{x \to 1} \frac{x^2 – 1}{x – 1} \) using a table of values.
▶️ Answer/Explanation
Step 1: Make a table
| x | f(x) |
|---|---|
| 0.9 | 1.9 |
| 0.99 | 1.99 |
| 0.999 | 1.999 |
| 1.001 | 2.001 |
| 1.01 | 2.01 |
| 1.1 | 2.1 |
Step 2: Estimate from table
As \( x \to 1 \), \( f(x) \) approaches 2 from both sides.
Conclusion: \( \lim_{x \to 1} \frac{x^2 – 1}{x – 1} = 2 \) (estimated from the table and graph would confirm this)
Example:
The table below gives values of \( f(x) \) near \( x = 3 \). Estimate \( \lim_{x \to 3} f(x) \).
| x | f(x) |
|---|---|
| 2.9 | 7.9 |
| 2.99 | 7.99 |
| 2.999 | 7.999 |
| 3.001 | 8.001 |
| 3.01 | 8.01 |
| 3.1 | 8.1 |
▶️ Answer/Explanation
Looking at the table, as \( x \) approaches 3 from both sides:
- From the left: \( f(x) \) approaches values near 8 (7.9 → 7.99 → 7.999)
- From the right: \( f(x) \) approaches values near 8 (8.001 → 8.01 → 8.1)
Since \( f(x) \) approaches 8 from both sides:
\( \lim_{x \to 3} f(x) = 8 \)
Using a graph:
Observe the \( y \)-values of \( f(x) \) as \( x \) approaches \( a \). The height the curve approaches is the limit.
Notation: If the graph suggest the same value, we write:
\( \lim_{x \to a} f(x) = L \)
Example:
Using the graph of the function y=f(x) shown in Figure, estimate the following limits.
▶️ Answer/Explanation
Solution:
- a. 0
- b. 2;
- c. does not exist
- d.−2
- e. 0
- f. does not exist
- g. 4
- h. 4
- i. 4
The Derivative as Gradient Function and Rate of Change
The Derivative as Gradient Function and Rate of Change
The derivative of a function \( f(x) \) measures how \( f(x) \) changes as \( x \) changes. It can be interpreted as:
- Gradient function: The derivative gives the gradient (slope) of the tangent line to the curve at each point \( x \).
- Rate of change: The derivative represents how fast the dependent variable changes with respect to the independent variable.
Forms of notation for the first derivative:
- \( \frac{dy}{dx} \) — derivative of \( y \) with respect to \( x \)
- \( f'(x) \) — derivative of the function \( f \)
- \( \frac{dV}{dr} \) — rate of change of volume \( V \) with radius \( r \)
- \( \frac{ds}{dt} \) — rate of change of displacement \( s \) with time \( t \)
Example:
The radius \( r \) of a balloon is increasing over time according to the equation \( r(t) = \sqrt{t^2 + 1} \) where \( r \) is in cm and \( t \) is in seconds.
Find the rate at which the surface area \( S \) of the balloon is increasing when \( t = 3 \).
(Surface area of a sphere: \( S = 4 \pi r^2 \))
▶️ Answer/Explanation
\( S = 4 \pi r^2 \)
\( \frac{dS}{dt} = 4 \pi \cdot 2r \frac{dr}{dt} = 8 \pi r \frac{dr}{dt} \)
\( r = (t^2 + 1)^{1/2} \)
\( \frac{dr}{dt} = \frac{1}{2} (t^2 + 1)^{-1/2} \cdot 2t = \frac{t}{\sqrt{t^2 + 1}} \)
Evaluate at \( t = 3 \)
\( r = \sqrt{3^2 + 1} = \sqrt{10} \)
\( \frac{dr}{dt} = \frac{3}{\sqrt{10}} \)
Compute \( \frac{dS}{dt} \)
\( \frac{dS}{dt} = 8 \pi \sqrt{10} \cdot \frac{3}{\sqrt{10}} = 8 \pi \cdot 3 = 24 \pi \)
Conclusion: The surface area is increasing at a rate of \( 24 \pi \ \text{cm}^2/\text{s} \) when \( t = 3 \).
Informal Understanding of the Gradient of a Curve as a Limit
Informal Understanding of the Gradient of a Curve as a Limit
The gradient of a curve at a point is the slope of the tangent line at that point. Since a curve doesn’t have a single gradient between two points, we estimate the gradient at a point by considering the gradient of a chord (secant) and letting the two points get infinitely close.
If we take two points on the curve \( (x, f(x)) \) and \( (x + h, f(x + h)) \), the gradient of the chord is:
\( \frac{f(x + h) – f(x)}{h} \)
As \( h \to 0 \), this approaches the gradient of the tangent:
\( \lim_{h \to 0} \frac{f(x + h) – f(x)}{h} \)
This limit gives the derivative at \( x \), which is the gradient of the curve at that point.
Example:
Estimate the gradient of \( f(x) = x^2 \) at \( x = 2 \) by computing the chord gradient for \( h = 0.1 \), \( 0.01 \), and \( 0.001 \).
▶️ Answer/Explanation
Gradient formula:
\( \frac{(2 + h)^2 – 2^2}{h} = \frac{(4 + 4h + h^2) – 4}{h} = \frac{4h + h^2}{h} = 4 + h \)
Calculate for each \( h \):
- \( h = 0.1 \): \( 4 + 0.1 = 4.1 \)
- \( h = 0.01 \): \( 4 + 0.01 = 4.01 \)
- \( h = 0.001 \): \( 4 + 0.001 = 4.001 \)
As \( h \to 0 \), the gradient approaches 4.
Conclusion: The gradient of the curve at \( x = 2 \) is 4.