Limits Revision Notes for Grade 12 NSC Matric Mathematics

Limits

What are limits?

Limits are a fundamental concept in calculus that help us understand how functions behave as we approach specific values. A limit describes the value that a function gets closer and closer to, even if it never actually reaches that value.

Think of limits as asking the question: "What happens to the function values as we get very close to a particular x-value?"

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The key idea is that we're interested in the behaviour of a function near a point, not necessarily at that exact point. This becomes especially important when the function is undefined at that point.

Understanding limits through examples

Let's examine a function to see how limits work in practice. Consider the rational function:

$y = \frac{x^2 + 4x - 12}{x + 6}$

We can factorise the numerator: $y = \frac{(x + 6)(x - 2)}{x + 6}$

When $x ≠ -6$ , we can cancel the common factor $(x + 6)$ , leaving us with $y = x - 2$ .

However, when $x = -6$ , the denominator becomes zero, making the function undefined at this point. But we can still investigate what happens as $x$ approaches $-6$ .

The table shows that as $x$ gets closer and closer to $-6$ from both sides, the $y$ -values approach $-8$ . This means the limit of the function as $x$ approaches $-6$ is $-8$ .

The graph shows a straight line with a "hole" at $x = -6$ , where the function would equal $-8$ if it were defined there.

Limit notation

We use special mathematical notation to express limits:

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$\lim_{x \to a} f(x) = L$

This reads: "The limit of $f(x)$ as $x$ approaches $a$ equals $L$ "

For our previous example, we write:

$\lim_{x \to -6} \frac{x^2 + 4x - 12}{x + 6} = -8$

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Worked Example: Basic Limit Notation

Question: Write the following using limit notation: as $x$ gets close to $1$ , the value of the function $y = x + 2$ approaches $3$ .

Solution: This is written as: $\lim_{x \to 1} (x + 2) = 3$

The diagram shows that when $x = 1$ , the function $y = x + 2$ equals $3$ .

One-sided limits

Sometimes a function approaches different values depending on whether we approach from the left or the right side.

Left-hand limit: $\lim_{x \to a^-} f(x)$ (approaches from the left)
Right-hand limit: $\lim_{x \to a^+} f(x)$ (approaches from the right)

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When both one-sided limits exist and are equal, the limit exists. When they're different, the limit does not exist.

This absolute value function demonstrates how limits can vary based on the direction of approach.

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Worked Example: Simple Limits

Question: Determine:

$\lim_{x \to 1} 10$
$\lim_{x \to 2} (x + 4)$

Solution:

Step 1: Simplify the expressions
We cannot simplify further and there are no terms to cancel.

Step 2: Calculate the limits

$\lim_{x \to 1} 10 = 10$ (constant function)
$\lim_{x \to 2} (x + 4) = 2 + 4 = 6$

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Worked Example: Rational Function Limits

Question: Determine: $\lim_{x \to 10} \frac{x^2 - 100}{x - 10}$

Solution:

Step 1: Simplify the expression
Factorise the numerator: $\frac{x^2 - 100}{x - 10} = \frac{(x + 10)(x - 10)}{x - 10}$

Step 2: Cancel common terms
$\frac{(x + 10)(x - 10)}{x - 10} = x + 10$

Step 3: Calculate the limit
$\lim_{x \to 10} \frac{x^2 - 100}{x - 10} = \lim_{x \to 10} (x + 10) = 10 + 10 = 20$

Gradient at a point

The concept of limits helps us find the gradient (slope) of a curve at any specific point. This connects to the idea of tangent lines.

Average gradient vs instantaneous gradient

The average gradient between two points on a curve is calculated using:

$\text{Average gradient} = \frac{y_Q - y_P}{x_Q - x_P} = \frac{f(a + h) - f(a)}{h}$

As point $Q$ moves closer to point $P$ , the secant line approaches the tangent line.

Gradient at a point formula

The gradient at a point (instantaneous gradient) is found using limits:

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$\text{Gradient at point P} = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

This formula is fundamental to differential calculus and forms the basis for finding derivatives.

This diagram shows how multiple secant lines approach the tangent line as the second point gets closer to the first.

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Worked Example: Finding Gradient at a Point

Question: Given $g(x) = 3x^2$ , determine the gradient of the curve at the point $x = -1$ .

Solution:

Step 1: Write the gradient formula
$\text{Gradient at a point} = \lim_{h \to 0} \frac{g(a + h) - g(a)}{h}$

Step 2: Determine $g(a + h)$ and $g(a)$
With $a = -1$ :

$g(a) = g(-1) = 3(-1)^2 = 3$
$g(a + h) = g(-1 + h) = 3(-1 + h)^2 = 3(1 - 2h + h^2) = 3 - 6h + 3h^2$

Step 3: Substitute and simplify
$\lim_{h \to 0} \frac{g(-1 + h) - g(-1)}{h} = \lim_{h \to 0} \frac{(3 - 6h + 3h^2) - 3}{h}$ $= \lim_{h \to 0} \frac{-6h + 3h^2}{h} = \lim_{h \to 0} \frac{h(-6 + 3h)}{h} = \lim_{h \to 0} (-6 + 3h) = -6$

Step 4: Final answer
The gradient of the curve $g(x) = 3x^2$ at $x = -1$ is $-6$ .

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Worked Example: More Complex Gradient Calculation

Question: Given $f(x) = 2x^2 - 5x$ , determine the gradient of the tangent to the curve at $x = 2$ .

Solution:

Step 1: Write the gradient formula
$\text{Gradient at a point} = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Step 2: Determine $f(a + h)$ and $f(a)$
With $a = 2$ :

$f(a) = f(2) = 2(2)^2 - 5(2) = 8 - 10 = -2$
$f(a + h) = f(2 + h) = 2(2 + h)^2 - 5(2 + h) = 2(4 + 4h + h^2) - 10 - 5h = 8 + 8h + 2h^2 - 10 - 5h = -2 + 3h + 2h^2$

Step 3: Substitute and simplify
$\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0} \frac{(-2 + 3h + 2h^2) - (-2)}{h}$ $= \lim_{h \to 0} \frac{3h + 2h^2}{h} = \lim_{h \to 0} \frac{h(3 + 2h)}{h} = \lim_{h \to 0} (3 + 2h) = 3$

Step 4: Final answer
The gradient of the tangent to the curve $f(x) = 2x^2 - 5x$ at $x = 2$ is $3$ .

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Remember that finding the gradient at a point using limits is essentially finding the derivative of the function at that point. This connection between limits and derivatives is one of the most important relationships in calculus.

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Key Exam Tips:

Always check if direct substitution works first - if it doesn't give $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , that's your answer
Factorise and cancel when you get indeterminate forms like $\frac{0}{0}$
Remember the gradient formula - it's the foundation for derivatives
One-sided limits may be different - check both sides
Show all working clearly - examiners want to see your method

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Key Points to Remember:

Limits describe what value a function approaches, not necessarily what it equals
Use the notation $\lim_{x \to a} f(x) = L$ to express limits mathematically
One-sided limits from left ( $x \to a^-$ ) and right ( $x \to a^+$ ) must be equal for the limit to exist
The gradient at a point is found using $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$
Secant lines approach the tangent line as points get closer together

Limits (Grade 12 NSC Matric Mathematics): Revision Notes

Limits

What are limits?

Understanding limits through examples

Limit notation

One-sided limits

Gradient at a point

Average gradient vs instantaneous gradient

Gradient at a point formula

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