What Is a Limit? Understanding Limits Intuitively

Limits are the foundation of calculus. Derivatives, integrals and continuity are all defined using limits. Yet many FSc, A Level and first-year university students learn to calculate limits without understanding what they mean. This guide builds the intuition first and then shows you how to evaluate limits confidently.

The Big Idea

A limit asks: what value does \( f(x) \) approach as \( x \) gets closer and closer to some number \( a \)? We write

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

and read it as "the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \)".

The key word is approaches. A limit describes the behaviour of a function near \( a \), not at \( a \). The function might not even be defined at \( a \), and the limit can still exist.

A simple illustration

Consider

$$ f(x) = \frac{x^2 - 1}{x - 1} $$

At \( x = 1 \) this gives \( \frac{0}{0} \), which is undefined. But look at nearby values:

• \( f(0.9) = \frac{0.81 - 1}{-0.1} = 1.9 \)
• \( f(0.99) = 1.99 \)
• \( f(1.01) = 2.01 \)
• \( f(1.1) = 2.1 \)

As \( x \) gets closer to 1 from either side, \( f(x) \) gets closer to 2. So

$$ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2 $$

even though \( f(1) \) does not exist. On a graph, this is a straight line \( y = x + 1 \) with a "hole" at \( (1, 2) \).

One-Sided Limits

Sometimes a function behaves differently depending on the side you approach from.

• Left-hand limit: \( \lim_{x \to a^-} f(x) \), approaching from values less than \( a \).
• Right-hand limit: \( \lim_{x \to a^+} f(x) \), approaching from values greater than \( a \).

The two-sided limit exists only if both one-sided limits exist and are equal:

$$ \lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L $$

Example: For \( f(x) = \frac{|x|}{x} \), the value is \(-1\) for all negative \( x \) and \( +1 \) for all positive \( x \). So the left-hand limit at 0 is \(-1\), the right-hand limit is \( 1 \), and \( \lim_{x \to 0} \frac{|x|}{x} \) does not exist.

Limits at Infinity

We can also ask what happens as \( x \) grows without bound. For example, as \( x \to \infty \), \( \frac{1}{x} \) gets smaller and smaller:

$$ \lim_{x \to \infty} \frac{1}{x} = 0 $$

This tells us the graph has a horizontal asymptote at \( y = 0 \). For rational functions, divide every term by the highest power of \( x \) in the denominator.

How to Evaluate Limits

Work through these methods in order:

1. Direct substitution. If plugging in \( a \) gives a real number, that is the limit (for polynomials and other continuous functions).
2. Factorise and cancel. If you get \( \frac{0}{0} \), factorise and cancel the common factor.
3. Rationalise. If square roots cause \( \frac{0}{0} \), multiply by the conjugate.
4. Use standard limits, such as

$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad (x \text{ in radians}) $$

5. Divide by the highest power for limits at infinity.

Note that \( \frac{0}{0} \) is called an indeterminate form. It does not mean the limit is 0 or undefined; it means you need more work.

Worked Example 1: Factorising

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

Solution:

• Step 1: Try substitution. \( \frac{9 - 9}{3 - 3} = \frac{0}{0} \), indeterminate.
• Step 2: Factorise the numerator.

$$ \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} $$

• Step 3: Cancel (allowed because \( x \ne 3 \) while approaching 3).

$$ = x + 3 $$

• Step 4: Substitute.

$$ \lim_{x \to 3} (x + 3) = 6 $$

Worked Example 2: Rationalising

Question: Evaluate \( \displaystyle \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \).

Solution:

• Step 1: Substitution gives \( \frac{2 - 2}{0} = \frac{0}{0} \).
• Step 2: Multiply by the conjugate.

$$ \frac{\sqrt{x + 4} - 2}{x} \times \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} = \frac{(x + 4) - 4}{x\left(\sqrt{x + 4} + 2\right)} $$

• Step 3: Simplify and cancel \( x \).

$$ = \frac{x}{x\left(\sqrt{x + 4} + 2\right)} = \frac{1}{\sqrt{x + 4} + 2} $$

• Step 4: Substitute \( x = 0 \).

$$ \frac{1}{\sqrt{4} + 2} = \frac{1}{4} $$

Worked Example 3: Limit at Infinity

Question: Evaluate \( \displaystyle \lim_{x \to \infty} \frac{3x^2 + 5x}{2x^2 - 7} \).

Solution:

• Step 1: Divide every term by \( x^2 \).

$$ \frac{3 + \frac{5}{x}}{2 - \frac{7}{x^2}} $$

• Step 2: As \( x \to \infty \), \( \frac{5}{x} \to 0 \) and \( \frac{7}{x^2} \to 0 \).

$$ \lim_{x \to \infty} \frac{3x^2 + 5x}{2x^2 - 7} = \frac{3}{2} $$

So the graph has a horizontal asymptote at \( y = \frac{3}{2} \).

Common Mistakes

• Writing \( \frac{0}{0} = 0 \) or stopping there. It is a signal to simplify further.
• Confusing \( \lim_{x \to a} f(x) \) with \( f(a) \). They are equal only for continuous functions.
• Ignoring one-sided limits for piecewise or absolute-value functions.
• Using degrees with \( \frac{\sin x}{x} \). The standard limit requires radians.
• Dropping the "lim" notation before the final substitution step, which costs marks in exams.

Key Takeaways

• A limit describes what \( f(x) \) approaches near \( a \), not its value at \( a \).
• A two-sided limit exists only when the left and right limits agree.
• \( \frac{0}{0} \) is indeterminate: factorise, rationalise or use a standard limit.
• For limits at infinity, divide by the highest power of \( x \) in the denominator.
• Limits are the building block for continuity, derivatives and integrals.

If limits still feel abstract or your calculus assignment is due soon, Amathify's experts and PhD scholars can walk you through step-by-step solutions. Reach us on WhatsApp or send your question using the "Ask a Question" form.