Limits and Continuity Revision Notes for VCE SSCE Mathematical Methods

Limits and Continuity

Understanding limits

A limit describes the value that a function approaches as the input approaches a particular point. This is a fundamental concept in calculus that helps us understand the behaviour of functions.

The notation $\lim_{x \to a} f(x) = p$ tells us that as $x$ gets closer and closer to the value $a$ , the function $f(x)$ approaches the value $p$ . We can make $f(x)$ as close as we like to $p$ by choosing values of $x$ sufficiently close to $a$ .

For many functions, particularly polynomial functions, finding the limit at a point is straightforward - we simply substitute the value into the function.

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Worked Example: Finding a Simple Limit

Find $\lim_{x \to 2} 3x^2$ .

Solution:

\lim_{x \to 2} 3x^2 = 3(2)^2 = 12

Explanation: As $x$ gets closer and closer to $2$ , the value of $3x^2$ gets closer and closer to $12$ .

Limits when functions are undefined

When a function is not defined at a particular value, we need a different approach. Often this involves algebraic manipulation, such as factoring and cancelling common terms.

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Worked Example: Limit with Undefined Function

For $f(x) = \frac{2x^2 - 5x + 2}{x - 2}$ , where $x \neq 2$ , find $\lim_{x \to 2} f(x)$ .

Solution:

Observe that:

f(x) = \frac{2x^2 - 5x + 2}{x - 2}

Hence $\lim_{x \to 2} f(x) = 3$ .

We can investigate this further by examining values of the function as $x$ approaches $2$ from both sides:

Table

From the table, we can see that as $x$ takes values closer to $2$ from either the left or the right, the function values approach $3$ .

chatImportant

The limit exists even though the function is not defined at $x = 2$ . This is a crucial distinction - the limit describes the behaviour near a point, not necessarily at the point itself.

Algebra of limits

When working with limits, we can use several important rules that make calculations easier. These rules assume that both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist.

Sum rule

\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)

The limit of the sum is the sum of the limits.

Multiple rule

\lim_{x \to a} k f(x) = k \lim_{x \to a} f(x)

where $k$ is a constant. We can take constant factors outside the limit.

Product rule

\lim_{x \to a} [f(x) \times g(x)] = \lim_{x \to a} f(x) \times \lim_{x \to a} g(x)

The limit of the product is the product of the limits.

Quotient rule

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}

provided $\lim_{x \to a} g(x) \neq 0$

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The limit of the quotient is the quotient of the limits, as long as the denominator's limit is not zero. This restriction is essential to avoid division by zero.

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Worked Example: Applying Limit Rules

Find the following limits:

a) $\lim_{h \to 0} (3h + 4)$

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

c) $\lim_{x \to 3} \frac{5x + 2}{x - 2}$

Solution:

\lim_{h \to 0} (3h + 4) = \lim_{h \to 0} (3h) + \lim_{h \to 0} (4)

\lim_{x \to 2} 4x(x + 2) = \lim_{x \to 2} (4x) \times \lim_{x \to 2} (x + 2)

\lim_{x \to 3} \frac{5x + 2}{x - 2} = \lim_{x \to 3} (5x + 2) \div \lim_{x \to 3} (x - 2)

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Worked Example: Factoring to Find Limits

Find the following limits by factoring first:

a) $\lim_{x \to 3} \frac{x^2 - 3x}{x - 3}$

b) $\lim_{x \to 2} \frac{x^2 - x - 2}{x - 2}$

c) $\lim_{x \to 3} \frac{x^2 - 7x + 10}{x^2 - 25}$

Solution:

\lim_{x \to 3} \frac{x^2 - 3x}{x - 3} = \lim_{x \to 3} \frac{x(x - 3)}{x - 3}

\lim_{x \to 2} \frac{x^2 - x - 2}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 1)}{x - 2}

\lim_{x \to 3} \frac{x^2 - 7x + 10}{x^2 - 25} = \lim_{x \to 3} \frac{(x - 2)(x - 5)}{(x + 5)(x - 5)}

Left and right limits

Sometimes it's useful to consider what happens when we approach a value from one side only. This is particularly important for piecewise-defined functions.

Right-hand limit

If $f(x)$ approaches the value $p$ as $x$ approaches $a$ from the right (from values greater than $a$ ), we write:

\lim_{x \to a^+} f(x) = p

Left-hand limit

If $f(x)$ approaches the value $p$ as $x$ approaches $a$ from the left (from values less than $a$ ), we write:

\lim_{x \to a^-} f(x) = p

Existence of limits

chatImportant

The limit $\lim_{x \to a} f(x)$ exists only if both the left-hand and right-hand limits exist and are equal. In that case, we can write $\lim_{x \to a} f(x) = p$ .

Piecewise-defined functions

For piecewise-defined functions, the limit may not exist at certain points, even though the function is defined there.

Consider the function:

f(x) = \begin{cases} x^3 & \text{if } 0 \leq x < 1 \\ 5 & \text{if } x = 1 \\ 6 & \text{if } 1 < x \leq 2 \end{cases}

From the graph, we can see that $\lim_{x \to 1} f(x)$ does not exist. However, if we approach $1$ from the left, $f(x)$ approaches $1$ , and if we approach from the right, $f(x)$ approaches $6$ . Also note that $f(1) = 5$ .

infoNote

This shows that the function value, the left-hand limit, and the right-hand limit can all be different. This is a key characteristic of discontinuous functions.

Asymptotic behaviour and the rectangular hyperbola

Limit notation is useful for describing how functions behave near vertical and horizontal asymptotes.

Consider the function $g(x) = \frac{1}{x}$ for $x \neq 0$ .

As $x$ approaches $0$ from the left, the function decreases without bound, while from the right it increases without bound. We write:

\lim_{x \to 0^-} g(x) = -\infty \text{ and } \lim_{x \to 0^+} g(x) = \infty

As $x$ increases towards positive infinity, the function values approach zero from above, and as $x$ decreases towards negative infinity, the values approach zero from below:

\lim_{x \to \infty} g(x) = 0^+ \text{ and } \lim_{x \to -\infty} g(x) = 0^-

infoNote

The notation $0^+$ indicates approaching zero from positive values (from above), while $0^-$ indicates approaching zero from negative values (from below).

Continuity at a point

Understanding continuity helps us identify where functions have breaks or gaps in their graphs.

Intuitive definition

A function is continuous at $x = a$ if we can draw the graph through the point $(a, f(a))$ without lifting our pen. If we cannot do this, there is a discontinuity at $x = a$ .

Formal definition

A function $f$ is continuous at the point $x = a$ if:

$f(x)$ is defined at $x = a$
$\lim_{x \to a} f(x) = f(a)$

chatImportant

Both conditions must be satisfied for the function to be continuous at that point. If either condition fails, the function is discontinuous at that point.

A function is discontinuous at a point if it is not continuous there.

A function is continuous everywhere if it is continuous for all real numbers. All polynomial functions are continuous everywhere.

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Worked Example: Identifying Discontinuities from Graphs

State the values of $x$ for which the functions shown below have a discontinuity:

Solution:

a) Discontinuity at $x = 1$ , as $f(1) = 3$ but $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^-} f(x) = 2$ .

b) Discontinuity at $x = -1$ , as $f(-1) = 2$ and $\lim_{x \to -1^-} f(x) = 2$ but $\lim_{x \to -1^+} f(x) = -\infty$ ,

and a discontinuity at $x = 1$ , as $f(1) = 2$ and $\lim_{x \to 1^-} f(x) = 2$ but $\lim_{x \to 1^+} f(x) = 3$ .

c) Discontinuity at $x = 1$ , as $f(1) = 1$ and $\lim_{x \to 1^-} f(x) = 1$ but $\lim_{x \to 1^+} f(x) = 2$ .

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Worked Example: Analyzing Continuity of Piecewise Functions

For each function, state the values of $x$ for which there is a discontinuity, and explain why using limits:

a) $f(x) = \begin{cases} 2x & \text{if } x \geq 0 \\ -2x + 1 & \text{if } x < 0 \end{cases}$

b) $f(x) = \begin{cases} x^2 & \text{if } x \geq 0 \\ -2x + 1 & \text{if } x < 0 \end{cases}$

c) $f(x) = \begin{cases} x & \text{if } x \leq -1 \\ x^2 & \text{if } -1 < x < 0 \\ -2x + 1 & \text{if } x \geq 0 \end{cases}$

d) $f(x) = \begin{cases} x^2 + 1 & \text{if } x \geq 0 \\ -2x + 1 & \text{if } x < 0 \end{cases}$

e) $f(x) = \begin{cases} x & \text{if } x \geq 0 \\ -2x & \text{if } x < 0 \end{cases}$

Solution:

a) $f(0) = 0$ but $\lim_{x \to 0^-} f(x) = 1$ , therefore there is a discontinuity at $x = 0$ .

b) $f(0) = 0$ but $\lim_{x \to 0^-} f(x) = 1$ , therefore there is a discontinuity at $x = 0$ .

c) $f(-1) = -1$ but $\lim_{x \to -1^+} f(x) = 1$ , therefore there is a discontinuity at $x = -1$ .

$f(0) = 1$ but $\lim_{x \to 0^-} f(x) = 0$ , therefore there is a discontinuity at $x = 0$ .

d) No discontinuity.

e) No discontinuity.

bookmarkSummary

Key Points to Remember:

The limit $\lim_{x \to a} f(x) = p$ means that $f(x)$ approaches $p$ as $x$ approaches $a$ .
For continuity at $x = a$ , two conditions must be met: the function must be defined at $a$ , and $\lim_{x \to a} f(x) = f(a)$ .
The algebra of limits allows us to split complex limits into simpler parts: sums, products, quotients, and constant multiples can all be separated.
Left and right limits may differ at a point, particularly for piecewise-defined functions. The overall limit exists only when both one-sided limits exist and are equal.
All polynomial functions are continuous everywhere, making them particularly well-behaved mathematical objects.

Limits and Continuity (VCE SSCE Mathematical Methods): Revision Notes