# Definition of Limits

The concept of a limit is essential to differential calculus. We will see that calculating limits is necessary for finding the gradient of a tangent to a curve at any point on the curve.

Consider the following table of values for $f(x)=x^2$ where $x$ is less than $3$ but increasing and getting closer and closer to $3$.

\begin{array}{|c|c|c|c|c|c|} \hline

x & 2 & 2.9 & 2.99 & 2.999 & 2.9999 \\ \hline

f(x)=x^2 & 4 & 8.41 & 8.9401 & 8.99401 & 8.9994001 \\ \hline

\end{array}

We say that as $x$ approaches $3$ from the left, $f(x)$ approaches $9$ from below.

\begin{array}{|c|c|c|c|c|c|} \hline

x & 4 & 3.1 & 3.01 & 3.001 & 3.0001 \\ \hline

f(x)=x^2 & 16 & 9.61 & 9.0601 & 9.006001 & 9.00060001 \\ \hline

\end{array}

In this case we say that as $x$ approaches $3$ from right, $f(x)$ approaches $9$ from above.

In summary, we can now say that as $x$ approaches $3$ from either direction, $f(x)$ approaches a limit of $9$, and write

$$ \large \lim_{x \rightarrow 3} x^2 = 9$$

## Definition of a Limit

If $f(x)$ can be made as close as we like to some real number $A$ by making $x$ sufficiently close to, but not equal to $a$, then we say that $f(x)$ has a limit of $A$ as $x$ approaches $a$, and we write

$$ \large \lim_{x \rightarrow a} f(x) = A$$

In this case, $f(x)$ is said to converge to $A$ as $x$ approaches $a$.

It is important to note that in defining the limit of $f$ as $x$ approaches $a$, $x$ does not reach $a$. The limit is defined for $x$ close to but not equal to $a$. Whether the function $f$ is defined or not at $x=a$ is not important to the limit of $f$ as $x$ approaches $a$. What is important is the behaviour of the function as $x$ gets very close to $a$.

For example, if $f(x)=\dfrac{x^2+x-2}{x-1}$ and we wish to find the limit as $x \rightarrow 0$, it is tempting for us to substitute $x=1$ into $f(x)$.

Not only do we get the meaningless value $\dfrac{0}{0}$, but also we destroy the basic limit method.

Observe that if $f(x)=\dfrac{x^2+x-2}{x-1} = \dfrac{(x-1)(x+2)}{x-1}$

then

\( \begin{align}

f(x)=

\begin{cases}

x+2 & \text{if } x \ne 1 \\

\text{undefined} & \text{if } x = 1 \\

\end{cases}

\end{align} \)

The graph of $y=f(x)$ is the straight line $y=x+2$ with the point $(1,3)$ missing, called a point of discontinuity of the function.

However, even though this point is missing, the limit of $f(x)$ as $x$ approaches $1$ does not exist. In particular, as $x\rightarrow 1$ from either direction, $y=f(x) \rightarrow 3$.

We write $\displaystyle \lim_{x \rightarrow 1} \dfrac{x^2+x-2}{x-1} = 3$ which reads:

Practically, we do not need to sketch the graph of the functions each time to determine limits, and most can be found algebraically.

### Example 1

Evaluate $\displaystyle \lim_{x \rightarrow 4} x^2$.

\( \begin{align} \displaystyle

\lim_{x \rightarrow 4} x^2 &= 4^2 \\

&= 16

\end{align} \)

### Example 2

Evaluate $\displaystyle \lim_{x \rightarrow 0} \dfrac{x^2+4x}{x}$.

\( \begin{align} \displaystyle

\lim_{x \rightarrow 0} \dfrac{x^2+4x}{x} &= \lim_{x \rightarrow 0} \dfrac{x(x+4)}{x} \\

&= \lim_{x \rightarrow 0} (x+4) \\

&= 0+4 \\

&= 4

\end{align} \)

### Example 3

Evaluate $\displaystyle \lim_{x \rightarrow 4} \dfrac{x^2-16}{x-4}$.

\( \begin{align} \displaystyle

\lim_{x \rightarrow 4} \dfrac{x^2-16}{x-4} &= \lim_{x \rightarrow 4} \dfrac{(x-4)(x+4)}{x-4} \\

&= \lim_{x \rightarrow 4} (x+4) \\

&= 4+4 \\

&= 8

\end{align} \)

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