Short and Sweet Calculus

2.1 The 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

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