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Introduction to Limits

A limit asks a simple question: where is a function heading as its input closes in on a value? It is the idea that makes calculus possible, and most of the time it is just careful substitution.

MYP 5Extended MathsNumber Sequences & Pre-CalculusCriteria A · C · D~10 min read

Written by Fayek · MYP student

The idea of a limit

Imagine walking towards a doorway. You get closer and closer, but the question that matters is not where you finally stop, it is which doorway you are aiming at. A limit works the same way: it describes the value a function approaches as \(x\) gets ever closer to some target, whether or not the function actually reaches it.

Limit
The value \(f(x)\) heads towards as \(x\) approaches a chosen number \(a\). We write \(\displaystyle \lim_{x \to a} f(x) = L\), read as "the limit of \(f(x)\) as \(x\) tends to \(a\) is \(L\)".

The power of this idea is that the function need not even be defined at the target. We can ask where it is heading near \(x = 2\) even if \(x = 2\) itself causes trouble, and that is exactly what we do next.

Limits at a finite value

For any function built from powers of \(x\) that behaves nicely, the first thing to try is direct substitution: put the target value straight in. If you get a sensible number, that is the limit.

Worked example

Evaluate \(\displaystyle \lim_{x \to 3} (x^2 + 1)\).

1
The function \(x^2 + 1\) is defined everywhere, so try substituting \(x = 3\) directly.
2
\((3)^2 + 1 = 9 + 1\).
\(\displaystyle \lim_{x \to 3} (x^2 + 1) = 10\).

Factor and cancel

Sometimes substitution gives \(\tfrac{0}{0}\), which is undefined and tells you nothing on its own. This is called an indeterminate form. It usually means a common factor is hiding on the top and bottom. Factor, cancel it, then substitute again.

Worked example

Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x - 2}\).

1
Substituting \(x = 2\) gives \(\dfrac{4 - 4}{2 - 2} = \dfrac{0}{0}\), which is indeterminate, so do not stop here.
2
Factor the top as a difference of two squares: \(x^2 - 4 = (x - 2)(x + 2)\).
3
Cancel the common factor \((x - 2)\): \(\dfrac{(x - 2)(x + 2)}{x - 2} = x + 2\).
4
Now substitute \(x = 2\) into \(x + 2\): \(2 + 2 = 4\).
\(\displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4\).

Zero over zero is a signal, not an answer

Getting \(\tfrac{0}{0}\) is your cue to factor and cancel, not to write "undefined". The function has a hole at \(x = 2\), yet it still heads cleanly towards 4 from both sides, and that heading is the limit.

Behaviour as x grows

We can also ask what happens as \(x\) grows without bound, written \(x \to \infty\). The trick for a fraction of polynomials is to divide every term by the highest power of \(x\) present. Any term of the form \(\tfrac{\text{number}}{x}\) or \(\tfrac{\text{number}}{x^2}\) then shrinks towards 0.

\[ \lim_{x \to \infty} \frac{1}{x} = 0 \qquad \lim_{x \to \infty} \frac{1}{x^2} = 0 \]
Worked example

Evaluate \(\displaystyle \lim_{x \to \infty} \frac{5x + 2}{2x - 1}\).

1
The highest power of \(x\) is \(x^1\), so divide every term on top and bottom by \(x\).
2
This gives \(\dfrac{5 + \tfrac{2}{x}}{2 - \tfrac{1}{x}}\).
3
As \(x \to \infty\), the terms \(\tfrac{2}{x}\) and \(\tfrac{1}{x}\) both head to 0, leaving \(\dfrac{5 + 0}{2 - 0}\).
\(\displaystyle \lim_{x \to \infty} \frac{5x + 2}{2x - 1} = \frac{5}{2}\).

Show the step that vanishes

When you write \(\tfrac{2}{x} \to 0\), say why: the top stays fixed while the bottom grows without limit. Making that reasoning explicit is what earns Criterion C (communicating) and Criterion D (applying maths in context), rather than just landing on \(\tfrac{5}{2}\).

Check yourself

1. Evaluate \(\displaystyle \lim_{x \to 4} (2x - 3)\). +

The function is defined everywhere, so substitute directly: \(2(4) - 3 = 8 - 3 = \mathbf{5}\).

2. Evaluate \(\displaystyle \lim_{x \to 3} \frac{x^2 - 9}{x - 3}\). +

Substituting gives \(\tfrac{0}{0}\), so factor first: \(x^2 - 9 = (x - 3)(x + 3)\). Cancel \((x - 3)\) to get \(x + 3\), then substitute \(x = 3\): \(3 + 3 = \mathbf{6}\).

3. Evaluate \(\displaystyle \lim_{x \to \infty} \frac{4x^2 + 1}{2x^2 + x}\). +

Divide every term by the highest power, \(x^2\): \(\dfrac{4 + \tfrac{1}{x^2}}{2 + \tfrac{1}{x}}\). As \(x \to \infty\), both \(\tfrac{1}{x^2}\) and \(\tfrac{1}{x}\) head to 0, leaving \(\dfrac{4}{2} = \mathbf{2}\).

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