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Consider the function $f$ defined by the equation

\[f(x)=\frac{4x^{2}-4}{2x-2}.\]

$f$ is defined for all values of $x$ except $x=1$, because 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

\[\lim_{x\to1}f(x)=4\]

or

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

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

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

or

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

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

\[g(x)=\begin{cases}

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

3 & \text{if }x=1

\end{cases}

\]

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))

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

therefore

\[\lim_{x\to1}g(x)=4.\]

(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.

### When Numerical Approach Fails

#### Read more on when computers may give false answers

#### Show Less

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.

### Uniqueness of a Limit

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

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 $L_{1}=L_{2}$.