Consider the function $f$ defined by the equation


$f$ is defined for all values of $x$ except $x=1$, because the substitution of $x=1$ in the expression for $f(x)$ yields the undefined fraction $\frac{0}{0}$. But because $4x^{2}-4=4(x^{2}-1)=4(x-1)(x+1)$, if $x\neq1$ we can simplify the fraction as

\[f(x)=\frac{4x^{2}-4}{2x-2}=\frac{4\cancel{(x-1)}(x+1)}{2\cancel{(x-1)}}=2x+2\qquad(\text{if }x\neq1)\]

So the graph of $f$ is the line $y=2x+2$ with one point removed, namely $(1,4)$. This point is shown as a hole in Figure 1.

Figure 1: Graph of $f(x)=(4x^{2}-4)/(2x-2)$

Now let’s investigate the values of $f$ when $x$ is close to 1 but not equal to 1. Let $x$ take on the values $0.9,0.95,0.99,0.999,0.9999$, and so on or take on the values $1.1,1.05,1.01,1.001,1.0001$, and so on. The corresponding values of $f$ are shown in the following table.

From this table and graph of $f$ shown in Figure 1, we see that as $x$ gets closer and closer to 1 (on either side of 1), but not equal to 1, $f(x)$ gets closer and closer to 4; the closer $x$ is to 1, the closer $f(x)$ is to 4. More specifically, we can make the values of $f(x)$ as close to 4 as we desire by taking $x$ close enough to 2. We express this by saying that “the limit of $f(x)$ as $x$ approaches $1$ is $4$” or simply “$f(x)$ approaches 4 as $x$ approaches 1,” or “$f(x)$ tends to $4$ as $x$ tends to $1$,” and express it symbolically as



\[f(x)\to4\quad\text{as}\quad x\to1 \]

In general

Definition 1 (Unofficial Definition): If we can make the values of $f(x)$ as close as we please to a number $L$ by taking $x$ sufficiently close (but not equal) to $a$, we say “the limit of $f(x)$ as $x$ approaches $a$ is $L$” and write

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


\[f(x)\to L\quad\text{as}\quad x\to a\]

According to the above definition, $x$ approaches $a$ but $x\neq a$, so the nonexistence or existence of $f(x)$ at $x=a$ or the value of $f(a)$ (if exists) has no bearing on the existence or on the value of ${\displaystyle \lim_{x\to a}f(x)}$. For example, if we define the function $g$ as

\frac{4x^{2}-4}{2x-2} & \text{if }x\neq1\\
3 & \text{if }x=1

then $g(x)$ and $f(x)=\frac{4x^{2}-4}{2x-2}$ defined at the beginning of this section are basically the same except when $x=1$ (see Figure 2(a) and (b))

\[g(x)=2x+2,\qquad(\text{if } x\neq1)\]



(a) Graph of $f(x)=\dfrac{4x^{2}-4}{2x-2}$ (b) Graph of $g(x)$ defined above

Figure 2: As we can see $f(x)=g(x)$ except when $x=1$, and ${\displaystyle \lim_{x\to1}f(x)=\lim_{x\to1}g(x)=4.}$

For now to evaluate the limits, we use numerical and graphical approaches.

Example 1

Evaluate ${\displaystyle \lim_{x\to-1.2}x^{2}}$


Let $x$ approach $-1.2$ from both sides

As the above table shows and we expect from the graph of $y=x^{2}$ (Figure 3), as $x$ approaches $-1.2$, $y$ approaches $(-1.2)^{2}=1.44$.

Figure 3: Graph of  $y=x^2$

Example 2

Evaluate ${\displaystyle \lim_{x\to2}\frac{x^{2}-5x+6}{x-2}}$.


Let's choose some values of $x$ close to 2 and calculate the corresponding values of $y=(x^{2}-5x+6)/(x-2)$. The results are tabulated in the following table.

It appears as if $y$ were approaching the limit $-1$. To verify this, we factor the numerator and simplify the fraction as

\[y=\frac{x^{2}-5x+6}{x-2}=\frac{(x-2)(x-3)}{x-2}=x-3\quad(\text{if }x\neq2)\]

From this, it appears that $y$ can be made as near $-1$ as we please by taking $x$ sufficiently close to 2 (See Figure 4). Hence –1 is the limit of $y$ as $x$ approaches 2.

Figure 4: Graph of  $y=\dfrac{x^{2}-5x+6}{x-2}$

Example 3

Evaluate ${\displaystyle \lim_{x\to0}\frac{\sin x}{x}}$.


Again we note that the function $y=\sin x/x$ is not defined at $x=0$, but we can construct the following table by choosing some values of $x$ close to $0$ and calculate the corresponding values of $\sin x/x$. Recall that $\sin x$ means the angle $x$ is measured in radians, so you need to set your calculator in radian mode. (See the Section on Angles).

The graph of $y=\sin x/x$ is shown in the following figure. From the above table and this figure, we may conclude

\[\lim_{x\to0}\frac{\sin x}{x}=1.\]

Figure 5: Graph of $y=\dfrac{\sin x}{x}$.


When Numerical Approach Fails 


Read more on when computers may give false answers

In the previous examples, we used a calculator/computer to numerically evaluate the values of the given function $f(x)$ for $x$ near the given point. However, in some cases, computers may give false results. Here is an example of such situations.

Example 4

Evaluate ${\displaystyle \lim_{x\to0}\frac{1-\sqrt{1+x^{6}}}{x^{6}}}$.


The function


is not defined when $x=0$. Let’s construct a table to list the values of $f(x)$ for several values of $x$ near $0$.


From this table, we may conclude that

\[ \lim_{x\to0}\frac{1-\sqrt{1+x^{6}}}{x^{6}}=-0.5\]

We might be tempted to choose some values of $x$ closer to $0$. Let’s try it out.

The values in the above table are given by (see the following figure).

Figure 6: Calculations with WolframAlpha

If you use a calculator, you might get a different value, but eventually, you will get $0$ if you make $x$ sufficiently close to zero. Do these calculations show us the limit is $0$ instead of $-0.5$? Let’s graph this function (see the following figure). Again it seems that the limit is $-0.5$. 

Figure 7:  Graph of $y=\dfrac{1-\sqrt{1+x^{6}}}{x^{6}}$  when $-3\leq x\leq 3$

Let’s zoom in and graph this function again for values of $x$ closer to $0$.

Figure 8:  Graph of $y=\dfrac{1-\sqrt{1+x^{6}}}{x^{6}}$ for $-0.2\leq x\leq 0.2$

What is the reason for this strange behavior? This behavior originates from the fact that calculators and computers retain only a fixed number of digits during a calculation. For example, if a calculator stores only 4 significant digits, then 0.34246 and 0.34254 are both stored as 0.3425. In this specific example, when $x$ is very close to $0$, $\sqrt{1+x^{6}}$ is so close to 1 that the computer cannot distinguish between them. This undesirable effect is called “loss of significance.” It almost always occurs when subtracting two nearly equal numbers. Ways to avoid loss of significance are studied in numerical methods.

Here, we can multiply both the numerator and the denominator by the conjugate of the numerator (See the Section on Rationalizing Binomial Denominators).

\frac{1-\sqrt{1+x^{6}}}{x^{6}} & =\frac{1-\sqrt{1+x^{6}}}{x^{6}}\frac{1+\sqrt{1+x^{6}}}{1+\sqrt{1+x^{6}}}\\
& =\frac{1-\left(\sqrt{1+x^{6}}\right)^{2}}{x^{6}\left(1+\sqrt{1+x^{6}}\right)}\\
& =\frac{1-(1+x^{6})}{x^{6}\left(1+\sqrt{1+x^{6}}\right)}\\
& =\frac{-\cancel{x^{6}}}{\cancel{x^{6}}(1+\sqrt{1+x^{6}})}\\
& =-\frac{1}{1+\sqrt{1+x^{6}}}

Note that

\[y=\frac{1-\sqrt{1+x^{6}}}{x^{6}}\qquad\text{and}\qquad y=-\frac{1}{1+\sqrt{1+x^{6}}}\]

are equal except when $x=0$ because the first one is not defined for $x=0$ but the second one is. The following table lists the values of $y=-\frac{1}{1+\sqrt{1+x^{6}}}$ for several points approaching $x=0$.

Calculations with and the graph of $y=-\frac{1}{1+\sqrt{1+x^{6}}}$ are shown below.

Figure 9



Uniqueness of a Limit

Now let’s consider the sign function $y=\text{sgn}(x)$ introduced in the Section on Piecewise-Defined Functions

Example 5

Let the sign function be defined by

\large \text{sgn}(x)=\left\{\begin{matrix} 1 & \text{if } x>0 \\ 0 & \text{if } x=0\\ -1 & \text{if } x<0 \end{matrix}\right. .

Find ${\displaystyle \lim_{x\to0}\text{sgn}(x)}$.


The graph of this function is shown in the following figure . When $x$ approaches $0$ through positive numbers, $\text{sgn}(x)$ approaches $1$ [although $\text{sgn}(x)=1$ for $x>0$, we still say $\text{sgn}(x)$ approaches $1$]. When $x$ approaches $0$ through negative numbers, $\text{sgn}(x)$ approaches $-1$. Therefore, there is no single number that $\text{sgn}(x)$ approaches as $x$ approaches $0$ from either side. Therefore $\lim_{x\to0}\text{sgn}(x)$ does not exist.

Figure 10: Graph of $y=\text{sgn}(x)$

In general, the limit of $f(x)$ as $x$ approaches a number $a$ (if exists) is unique, because for all $x$ near $a$, $f(x)$ cannot be near two different numbers at the same time. 

Theorem 1 (Uniqueness of a limit): If ${\displaystyle {\lim_{x\to a}f(x)=L_{1}}}$ and ${\displaystyle {\lim_{x\to a}f(x)=L_{2}}}$ then $L_{1}=L_{2}$.