The idea of a limit

What you will be able to do

• read a limit from a table of values and from a graph;
• find a left-hand limit and a right-hand limit, and say when the limit does not exist;
• use the rules for the limit of a sum, a product and a quotient.

A hole you can still walk towards

A road has a hole at one spot. You can walk right up to the hole from either side, and you can say exactly how high the road would be there. A function can do the same. Take

At $x = 1$ the top and the bottom are both $0$, so $f(1)$ has no meaning. Now look at values near $1$.

The values head for $2$ from both sides. We say that the limit of $f(x)$ as $x$ tends to $1$ is $2$, and we write

The reason is simple. For every $x \ne 1$ we can cancel the common factor: $\dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1} = x + 1$. The graph is the line $y = x + 1$ with one point missing.

What the word limit means

$\displaystyle\lim_{x \to a} f(x) = L$ means that $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$, with $x \ne a$. The value $f(a)$ plays no part. It may be different from $L$, or it may not exist.

Books at university write this with two small numbers, $\varepsilon$ and $\delta$. School and entrance exams, and the first year of A-Level and IB, use the idea above.

Left and right

We can come to $a$ from the left or from the right.

• The left-hand limit, written $\displaystyle\lim_{x \to a^-} f(x)$, uses values of $x$ a little below $a$.
• The right-hand limit, written $\displaystyle\lim_{x \to a^+} f(x)$, uses values of $x$ a little above $a$.

The limit exists exactly when both one-sided limits exist and are equal. In the example above both are $2$. You will use the same idea again in the Class 12 chapter on continuity.

When a limit does not exist

There are three usual ways.

1. A jump. For $f(x) = \dfrac{|x|}{x}$ the left-hand limit at $0$ is $-1$ and the right-hand limit is $1$. They differ, so there is no limit.
2. Growing without bound. For $f(x) = \dfrac{1}{x}$ the values grow large and positive on the right and large and negative on the left. For $f(x) = \dfrac{1}{x^2}$ they grow large and positive on both sides. Books then write $\displaystyle\lim_{x \to 0} \dfrac{1}{x^2} = \infty$. This describes how the function behaves. As a number, the limit still does not exist.
3. Swinging. For $f(x) = \sin\dfrac{1}{x}$ the values swing between $-1$ and $1$ faster and faster near $0$, and settle on nothing.

The rules for limits

Suppose $\displaystyle\lim_{x \to a} f(x) = A$ and $\displaystyle\lim_{x \to a} g(x) = B$, where $A$ and $B$ are numbers. Then:

• the limit of $f + g$ is $A + B$, and the limit of $f - g$ is $A - B$;
• the limit of $cf$ is $cA$, for a constant $c$;
• the limit of $f \cdot g$ is $AB$;
• the limit of $\dfrac{f}{g}$ is $\dfrac{A}{B}$, if $B \ne 0$;
• the limit of $f(x)^n$ is $A^n$, for a whole number $n$.

Two facts follow. The limit of a constant is that constant, and the limit of $x$ as $x \to a$ is $a$. So for a polynomial, the limit at $a$ is its value at $a$. The same holds for a fraction of polynomials, as long as the bottom is not $0$ at $a$.

A worked example

Find $\displaystyle\lim_{x \to 2} (3x^2 - 5x + 1)$ and $\displaystyle\lim_{x \to 3} \dfrac{x^2 + 1}{x - 1}$.

1. The polynomial. Substitute $x = 2$: $3 \times 4 - 10 + 1 = 3$.
2. The fraction. At $x = 3$ the bottom is $2$, which is not zero. So the limit is $\dfrac{9 + 1}{3 - 1} = 5$.
3. A warning. Had the bottom been $0$, we would meet $\dfrac{0}{0}$ or $\dfrac{\text{number}}{0}$ and need more work. The next lesson shows how.

Practise with new numbers

Try it in Python

The program prints values of $f(x) = \dfrac{x^2 - 1}{x - 1}$ closer and closer to $1$ on both sides. It then looks at two functions that have no limit at $0$.

Common mistakes

• Using $f(a)$. A limit looks at the values near $a$, not at $a$. A function can have a limit where it has no value.
• Looking at one side only. Check both sides. A jump hides if you look from one side.
• Reading $\dfrac{0}{0}$ as no limit. It means that the work is not finished. Remove the common factor first.
• Treating $\infty$ as a number. A limit "equal to $\infty$" describes growth without bound. You cannot add or subtract it like a number.

Remember these

• $\displaystyle\lim_{x \to a} f(x) = L$ means that $f(x)$ gets as close to $L$ as we like when $x$ is close to $a$, with $x \ne a$.
• The limit exists when the left-hand limit equals the right-hand limit.
• A jump, growth without bound and swinging are the usual reasons for no limit.
• For a polynomial, the limit at $a$ is the value at $a$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. For $g(x) = \dfrac{x^2 - 9}{x - 3}$ the values at $x = 2.9,\ 2.99,\ 3.01,\ 3.1$ are $5.9,\ 5.99,\ 6.01,\ 6.1$. What is $\displaystyle\lim_{x \to 3} g(x)$?

2. Does $\displaystyle\lim_{x \to 0} \dfrac{|x|}{x}$ exist?

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

4. Let $f(x) = x + 1$ for $x \ne 2$ and $f(2) = 10$. What is $\displaystyle\lim_{x \to 2} f(x)$?

5. Does $\displaystyle\lim_{x \to 0} \dfrac{1}{x}$ exist?

References

• OpenStax, Calculus Volume 1, Chapter 2, sections 2.2 and 2.3 on the limit of a function and the limit laws.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the sections on intuitive idea of limits and algebra of limits.