Short and Sweet Calculus

## 2.1The Concept of a Limit

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 2.1.

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 2.1, 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

2.1. (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 [fig:Ch4-concept-limit-2]) $g(x)=2x+2,\qquad(\text{if }x\neq1)$ therefore $\lim_{x\to1}g(x)=4.$

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