Limits and continuity

Mathematics · First-year university mathematics · about 10 min

A limit describes the value a function approaches near a point, even if the function is not defined there. Continuity asks a further question: does the function actually take that limiting value? These ideas help you recognise breaks in graphs and give a precise foundation for calculus.

The key idea

A function is continuous at $x=a$ precisely when its limit exists and equals its value there:

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

Here $a$ is the input being approached, $f(a)$ is the actual function value, and the limit describes nearby behaviour.

Draw one Cartesian plot with the horizontal x-axis from 0 to 4 and vertical y-axis from 0 to 4 at equal scales. Draw the straight graph y=x from (0,0) to (4,4). Mark point P=(2,2) with a small solid dot. Draw a vertical dotted guide from P to x=2 and a horizontal dotted guide from P to y=2. Add two small arrows following the line, one from (1.2,1.2) towards (1.9,1.9) and one from (2.8,2.8) towards (2.1,2.1). The two arrows must both point towards P. Label the x-coordinate of P as a and the y-coordinate as L. Do not include tick numbers. — Matching approaches from the left and right produce a two-sided limit.

Try it yourself

Change the rational function and watch the graph approach its asymptotes and skip its holes.

Limits from both sides

Approach an input from smaller values to find the left-hand limit, $\lim_{x\to a^-}f(x)$ . Approach from larger values to find the right-hand limit, $\lim_{x\to a^+}f(x)$ . A finite two-sided limit exists only if these are equal. This distinction matters most for piecewise functions: a graph can approach two different heights at the same input. The value plotted exactly at $x=a$ does not decide whether the limit exists. Read the curve on either side first, then check any filled dot at the input.

Finding limits algebraically

Try direct substitution when the formula is continuous at the input. If it produces $0/0$ , that is an indeterminate form, not an answer. Factor and cancel a common factor where $x\ne a$ , or rationalise an expression containing a square root. For example, $\frac{x^2-4}{x-2}=x+2$ for $x\ne2$ , so its limit at $2$ is $4$ . Cancelling is valid for finding the nearby behaviour, but it does not automatically define the function at the missing input. When simplifying, keep track of the original domain.

Continuity and removable holes

To test continuity at $a$ , check three things: $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the two are equal. If the limit exists but $f(a)$ is missing or different, the discontinuity is removable: redefining one value repairs it. If left-hand and right-hand limits disagree, the break is a jump and changing only $f(a)$ cannot fix it. A continuous curve has no interruption at the point in question, but an isolated filled dot at another height changes continuity without changing the limit.

Draw Cartesian axes with x from 0 to 3 and y from 0 to 4, using equal linear scales. Draw the line y=x+1 over the displayed x interval, but omit the point (1,2) and mark it with a hollow circle. At the same x=1, draw a solid dot at (1,3). Draw a thin vertical dotted guide from x=1 on the horizontal axis to y=3. Label the hollow circle with its coordinate (1,2) and the solid point with f(1)=3. No other points or lines. — A single incorrect function value creates a removable discontinuity.

Worked example: Repairing a hole in a function

The function is defined by

f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\ne3,\\ k,&x=3.\end{cases}

Find $\lim_{x\to3}f(x)$ . Determine the value of $k$ that makes $f$ continuous at $x=3$ , and decide whether $k=5$ would be continuous.

Show the solution

Step 1: Simplify near the point. For $x\ne3$ , factor the numerator:

f(x)=\frac{(x-3)(x+3)}{x-3}=x+3.

This equality applies near $3$ , even though the original quotient is not used at $x=3$ .

Step 2: Find the limit.

\lim_{x\to3}f(x)=\lim_{x\to3}(x+3)=3+3=6.

Step 3: Apply continuity. At $x=3$ , the definition gives $f(3)=k$ . We need $f(3)=6$ , so $k=6$ .

Step 4: Check the alternative. If $k=5$ , then $f(3)=5$ but the limit is still $6$ . The graph has a removable discontinuity.

Answer: $\boldsymbol{\lim_{x\to3}f(x)=6}$ and $\boldsymbol{k=6}$ for continuity. $k=5$ is not continuous.

Common mistakes

Calling $0/0$ the limit.

Treat $0/0$ as a signal to simplify or use another limit method.

Looking only at the filled point.

Inspect values approaching from the left and the right before checking $f(a)$ .

Cancelling a factor and forgetting the hole.

State the condition $x\ne a$ and check the original definition at $a$ .

Assuming matching one-sided limits imply continuity.

Also check that $f(a)$ is defined and equals their common limit.

Check your understanding

1. Find $\lim_{x\to2}\frac{x^2-4}{x-2}$ .

2. Let $g(x)=2x+1$ for $x<1$ and $g(x)=x+4$ for $x\ge1$ . What is $\lim_{x\to1}g(x)$ ?

3. For $h(x)=\frac{x^2-1}{x-1}$ when $x\ne1$ and $h(1)=k$ , which $k$ makes $h$ continuous at $1$ ?

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