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Year 11 Methods (Unit 1 & 2) Differentiation Of Polynomials

Limits And Continuity

20 practice questions 1 video lesson Theory + worked examples

Learn limits and continuity for Queensland Year 11 Mathematical Methods (QCAA). A limit describes the value a function approaches as x nears a point, while continuity asks whether a graph can be drawn without lifting your pen.

You will learn to evaluate limits by direct substitution, resolve zero-over-zero limits by factoring, handle one-sided limits, and test whether a piecewise function is continuous — the basis for a derivative found from first principles.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), a limit \(\lim_{x\to a}f(x)\) is the value \(f(x)\) heads toward as \(x\) approaches \(a\). This page covers limits by direct substitution, removable \(\tfrac00\) limits by factor-and-cancel, one-sided limits and when a limit does not exist, and informal continuity at a point and of piecewise functions.

The limit \(\lim_{x\to a}f(x)=L\) means \(f(x)\) gets as close as we like to \(L\) as \(x\) approaches \(a\) from both sides. If the function is defined and unbroken there, you find it by direct substitution.

One-sided limits \(\lim_{x\to a^-}f(x)\) and \(\lim_{x\to a^+}f(x)\) approach from the left and right. The two-sided limit exists only when they are equal; at a jump they differ, so the limit does not exist (DNE).

A function is continuous at \(x=a\) if its graph has no break there: the limit exists and equals the function value, \(\lim_{x\to a}f(x)=f(a)\). A hole, jump or gap makes it discontinuous.

Substitute first. If substituting gives a number, that is the limit. If it gives \(\tfrac00\), factorise and cancel, then substitute again.
A removable discontinuity (a hole)Straight line y equals x plus 2 with a hollow point at x=2; the limit is 4 even though the function has a hole there. x y
A hole at \(x=2\): the limit is \(4\) even though the point itself is missing.
A jump discontinuityTwo branches that do not meet at x=1; the left limit is 2 and the right value is 3, so the two-sided limit does not exist. x y
A jump at \(x=1\): left limit \(2\), right value \(3\), so \(\lim_{x\to1}f(x)\) DNE.

Direct substitution (when \(f\) is defined and unbroken at \(a\)):

\[\lim_{x\to a}f(x)=f(a)\]
limx→af(x)=f(a)

Two-sided limit exists only when the one-sided limits agree:

\[\lim_{x\to a}f(x)=L\iff \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L\]
limx→a-f=limx→a+f
Continuity at \(a\): \(\lim_{x\to a}f(x)=f(a)\) — the limit exists and matches the actual value, so the graph joins up with no hole or jump.

Evaluating a limit (and checking continuity)

  1. Substitute \(x=a\). A defined value is the limit; \(\tfrac{\text{nonzero}}0\) means it grows without bound (DNE).
  2. If \(\tfrac00\): factorise numerator and denominator, cancel the common factor, then substitute again.
  3. Piecewise or a jump: take the left and right limits; if they differ the limit does not exist.
  4. Continuity: the function is continuous at \(a\) when the limit exists and equals \(f(a)\).
Example 1 — Direct substitution
Evaluate \(\displaystyle\lim_{x\to3}\,(x^2-2x+1)\).
Solution

The function is a polynomial (unbroken everywhere), so substitute \(x=3\):

\(\lim_{x\to3}(x^2-2x+1)\)\(=\)\((3)^2-2(3)+1\)
\(=\)\(9-6+1\)
\(=\)\(4\)

\(\displaystyle\lim_{x\to3}(x^2-2x+1)=4\).

Direct substitution into a polynomialParabola y equals x squared minus 2x plus 1 with the point at x=3, y=4 marked; the limit equals the value there. x y
limx→3(x2-2x+1)=4
Example 2 — A removable (0/0) limit
Evaluate \(\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}\).
Solution

Substitute \(x=2\) to test:

\(\dfrac{(2)^2-4}{2-2}\)\(=\)\(\dfrac{0}{0}\)

\(\tfrac00\) is indeterminate — factorise and cancel.

Factorise the numerator and cancel \((x-2)\):

\(\dfrac{x^2-4}{x-2}\)\(=\)\(\dfrac{(x-2)(x+2)}{x-2}\)
\(=\)\(x+2\)

Now substitute \(x=2\):

\(\lim_{x\to2}(x+2)\)\(=\)\(2+2\)
\(=\)\(4\)

\(\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}=4\).

Removable limit of (x squared minus 4) over (x minus 2)After cancelling, the graph is the line y equals x plus 2 with a hollow point at x=2; the limit is 4. x y
limx→2x2-4x-2=4
Example 3 — One-sided limits and DNE
For \(f(x)=\begin{cases}x+1,&x<1\\4-x,&x\ge1\end{cases}\) decide whether \(\displaystyle\lim_{x\to1}f(x)\) exists.
Solution

Left limit — use the \(x<1\) branch \(x+1\):

\(\lim_{x\to1^-}f(x)\)\(=\)\(1+1\)
\(=\)\(2\)

Right limit — use the \(x\ge1\) branch \(4-x\):

\(\lim_{x\to1^+}f(x)\)\(=\)\(4-1\)
\(=\)\(3\)

The one-sided limits are \(2\) and \(3\): they are not equal.

The one-sided limits differ (\(2\neq3\)), so \(\displaystyle\lim_{x\to1}f(x)\) does not exist.

One-sided limits at a jumpLeft branch approaches 2 and right branch takes value 3 at x=1; the left and right limits differ so the limit does not exist. x y
limx→1f(x) DNE
Example 4 — Making a piecewise function continuous
Find \(k\) so that \(f(x)=\begin{cases}2x+1,&x\le1\\x^2+k,&x>1\end{cases}\) is continuous at \(x=1\).
Solution

Left piece value at \(x=1\):

\(f(1)\)\(=\)\(2(1)+1\)
\(=\)\(3\)

Right-hand limit as \(x\to1^+\) (branch \(x^2+k\)):

\(\lim_{x\to1^+}f(x)\)\(=\)\((1)^2+k\)
\(=\)\(1+k\)

For continuity the two must be equal:

\(1+k\)\(=\)\(3\)
\(k\)\(=\)\(2\)

\(k=2\) makes \(f\) continuous at \(x=1\) (both pieces give \(3\)).

Making a piecewise function continuousWith k equal to 2 the two branches meet at x=1, y=3, so the piecewise function is continuous there. x y
k=2

Common pitfalls

Stopping at \(\tfrac00\). \(\tfrac00\) is not the answer — it signals a common factor to cancel. Factorise, cancel, then substitute again.
Assuming every limit exists. At a jump the left and right limits differ, so the two-sided limit does not exist.
Confusing the value with the limit. A function can have a hole: the limit exists even though \(f(a)\) is undefined or different there.
Ignoring which branch applies. For a one-sided limit, use the piece that covers that side of \(a\), not the other.

Frequently asked questions

How do you evaluate a limit?

Substitute the value first. If you get a number, that is the limit; if you get \(\tfrac00\), factorise and cancel, then substitute again.

What does it mean for a limit to not exist?

The function does not approach a single value — usually the left and right one-sided limits differ (a jump), or the function grows without bound.

What is a removable discontinuity?

A hole in the graph, from a factor that cancels (like \(\tfrac{x^2-4}{x-2}\)). The limit exists even though the point itself is missing.

When is a function continuous at a point?

When \(\lim_{x\to a}f(x)=f(a)\): the limit exists and equals the actual value, so the graph has no hole, jump or gap there.

How do you make a piecewise function continuous?

Set the two pieces equal at the join: make the value of one branch equal the limit of the other, then solve for the unknown.

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