Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Using Algebra to Evaluate Limits

In the next three sections, we consider evaluating limits at a point $a$ to which Theorem 3.14, does not apply; that is, limits of functions that are not continuous at $a\text{.}$

What happens when we simplify algebraic expressions? For example, to simplify

${\displaystyle \frac{x^2-4}{x-2}\text{,}}$

we would follow this procedure:

${\displaystyle \frac{x^2-4}{x-2}=\frac{\left(x-2\right)\left(x+2\right)}{x-2}=x+2;\text{thus}\frac{x^2-4}{x-2}=x+2\text{.}}$

When we are evaluating functions, this answer is not quite correct. What is correct in this equality is that

${\displaystyle \frac{x^2-4}{x-2}=x+2\text{is true (holds) for all }x\text{ except }2\text{.}}$

The function ${\displaystyle f\left(x\right)=\frac{x^2-4}{x-2}}$ is not defined at $2$, whereas the function $g\left(x\right)=x+2$ is defined for all $x\text{.}$

These two functions are not equal, because two functions $f$ and $g$ are equal only if they have the same domain and if $f\left(x\right)=g\left(x\right)$ for all $x$ in their domains.

The functions ${\displaystyle f\left(x\right)=\frac{x^2-4}{x-2}}$ and $g\left(x\right)=x+2$ do not have the same domain.

Figures 3.16 and 3.17, below, show the graphs of ${\displaystyle f\left(x\right)=\frac{x^2-4}{x-2}}$ and $g\left(x\right)=x+2$, respectively. As you can see, while the graph of $f$ has a hole because it is not defined at $2$, the graph of $g$ is continuous.

To evaluate the limit, we must consider only values of $f\left(x\right)$ for $x$ around $2\text{.}$ For these $x$s, we have $f\left(x\right)=g\left(x\right)$; that is, $f\left(x\right)=g\left(x\right)$ for $x\to a\text{.}$ Moreover, from the graphs of these functions, we see that indeed

$\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }g\left(x\right)\text{.}$

Figure 3.16.  Function ${\displaystyle f\left(x\right)=\frac{x^2-4}{x-2}}$

Figure 3.17. Function $g\left(x\right)=x+2$

Theorem 3.14 does not apply when we want to evaluate the limit $\underset{x\to 2}{\lim }f\left(x\right)$, because $f$ is not continuous at $2$, but it does apply to the limit $\underset{x\to 2}{\lim }g\left(x\right)\text{.}$ [Why?] Hence, we conclude that

$\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }g\left(x\right)=\underset{x\to 2}{\lim }x+2=4\text{.}$

In summary, we may be able to evaluate the limit $\underset{x\to a}{\lim }f\left(x\right)$ of a function $f$ which is not continuous at $a$, if we can obtain, through an algebraic manipulation of $f$, a function $g$, such that

$f\left(x\right)=g\left(x\right)\text{for all}x\text{around}a\text{,}\text{and}\underset{x\to a}{\lim }g\left(x\right)\text{exists.}$

Thus

$\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)\text{.}$

This conclusion is possible because of the following proposition.

Proposition 3.16.

1. If $f\left(x\right)=g\left(x\right)$ for all $x$ close to $a$ from the right, but not at $a$, then

   $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{+}}{\lim }g\left(x\right)\text{.}$

2. If $f\left(x\right)=g\left(x\right)$ for all $x$ close to $a$ from the left, but not at $a$, then

   $\underset{x\to a^{-}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }g\left(x\right)\text{.}$

3. If $f\left(x\right)=g\left(x\right)$ for all $x$ close to $a$, but not at $a$, then

   $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)\text{.}$

To use this approach in the evaluation of limits, we must learn which algebraic manipulations we can try for which functions. Here is where our algebraic skills will pay off. In the following examples, we try to cover the most common cases; you may be able to find other examples on your own.

Example 3.34. Rational functions.

Consider the limit

${\displaystyle \underset{x\to 3}{\lim }\frac{x^2+x-12}{3x^2-7x-6}\text{.}}$

The rational function

${\displaystyle f\left(x\right)=\frac{x^2+x-12}{3x^2-7x-6}}$

is not continuous at $3\text{.}$

If we evaluate the polynomials $p\left(x\right)=x^2+x-12$ and $q\left(x\right)=3x^2-7x-6$ at $3$, we get $0$ (try it).

There is a theorem that says that if $p\left(3\right)=0$, then $p$ can be factored as a product of the form $\left(x-3\right)r\left(x\right)$, where $r\left(x\right)$ is a polynomial. The same is true for $q\left(x\right)$: there is a polynomial $s\left(x\right)$ such that $q\left(x\right)=\left(x-3\right)s\left(x\right)\text{.}$ As we indicated in Unit 1,

$p\left(x\right)=\left(x-3\right)\left(x+4\right)\text{and}q\left(x\right)=\left(x-3\right)\left(3x+2\right)\text{.}$

Hence,

${\displaystyle \frac{x^2+x-12}{3x^2-7x-6}=\frac{\left(x-3\right)\left(x+4\right)}{\left(x-3\right)\left(3x+2\right)}=\frac{x+4}{3x+2}}$

for all $x$ around $3\text{.}$

The function

${\displaystyle g\left(x\right)=\frac{x+4}{3x+2}}$

is continuous at $3$, and we conclude that

${\displaystyle \underset{x\to 3}{\lim }\frac{x^2+x-12}{3x^2-7x-6}=\underset{x\to 3}{\lim }\frac{x+4}{3x+2}=\frac{7}{11}\text{.}}$

Note: If you do not remember how to factor a polynomial, but you know one of the factors, you can always use long division to find the other factor, as is shown below (see also Unit 1).

$\begin{array}{r}3x+2\\ x-<!--\; -\; -->3\overline{)\phantom{-<!--\; -\; -->}3x^2-<!--\; -\; -->7x-<!--\; -\; -->6}\\ \underset{\_<!--\; \_\; -->}{-<!--\; -\; -->3x^2+9x}\phantom{-<!--\; -\; -->}\phantom{6}\\ \phantom{-<!--\; -\; -->3x^2}\phantom{-<!--\; -\; -->}2x-<!--\; -\; -->6\\ \phantom{-<!--\; -\; -->3x^2}\underset{\_<!--\; \_\; -->}{-<!--\; -\; -->2x+6}\\ \phantom{-<!--\; -\; -->3x^2}\phantom{-<!--\; -\; -->}\phantom{2x}\phantom{-<!--\; -\; -->}0\end{array}$

Hence, $3x^2-7x-6=\left(x-3\right)\left(3x+2\right)\text{.}$

Warning: It is incorrect to write

${\displaystyle \underset{x\to 3}{\lim }\frac{x^2+x-12}{3x^2-7x-6}=\frac{x+4}{3x+2}=\frac{7}{11}\text{.}}$

This simply does not make sense: it is not true that

${\displaystyle \underset{x\to 3}{\lim }\frac{x^2+x-12}{3x^2-7x-6}=\frac{x+4}{3x+2}\text{.}}$

A limit is not equal to an algebraic expression, and the equality

${\displaystyle \frac{x+4}{3x+2}=\frac{7}{11}}$

is even worse: it is an equation and is not what we mean or want to mean.

Example 3.35. Algebraic operations.

The function in the limit

${\displaystyle \underset{h\to 0}{\lim }\frac{{\left(4+h\right)}^2-16}{h^2+h}}$

is also rational and discontinuous at $0\text{.}$ To simplify it, we do the algebraic operations and factorization as follows:

${\displaystyle \frac{{\left(4+h\right)}^2-16}{h^2+h}=\frac{16+8h+h^2-16}{h\left(h+1\right)}=\frac{8h+h^2}{h\left(h+1\right)}=\frac{8+h}{h+1}\text{.}}$

This is true for all $h\ne 0\text{.}$ We arrive at a continuous function at $0$, namely

${\displaystyle g\left(h\right)=\frac{8+h}{h+1}\text{.}}$

Hence,

${\displaystyle \underset{h\to 0}{\lim }\frac{{\left(4+h\right)}^2-16}{h^2+h}=\underset{h\to 0}{\lim }\frac{8+h}{h+1}=8\text{.}}$

Example 3.36. Fractions.

To simplify the function in the limit

${\displaystyle \underset{t\to 0}{\lim }\frac{1}{t}-\frac{1}{t^2+t}\text{,}}$

we must perform the operations of the indicated fractions.

${\displaystyle \frac{1}{t}-\frac{1}{t^2+t}=\frac{t^2+t-t}{t\left(t^2+t\right)}=\frac{t^2}{t^2\left(t+1\right)}=\frac{1}{t+1}\text{for }t\ne 0\text{.}}$

We again arrive at a continuous function at $0\text{.}$

${\displaystyle \underset{t\to 0}{\lim }\frac{1}{t}-\frac{1}{t^2+t}=\underset{t\to 0}{\lim }\frac{1}{t+1}=1\text{.}}$

Example 3.37. Rationalization.

The function in the limit

${\displaystyle \underset{h\to 0}{\lim }\frac{\sqrt{9+h}-3}{h}}$

is not continuous at $0$ and it involves square roots. So, we rationalize the function as follows:

${\displaystyle \left(\frac{\sqrt{9+h}-3}{h}\right)\left(\frac{\sqrt{9+h}+3}{\sqrt{9+h}+3}\right)=\frac{1}{\sqrt{9+h}+3}\text{for }h\ne 0\text{.}}$

We can apply Theorem 3.14 to conclude that

${\displaystyle \underset{h\to 0}{\lim }\frac{\sqrt{9+h}-3}{h}=\underset{h\to 0}{\lim }\frac{1}{\sqrt{9+h}+3}=\frac{1}{6}\text{.}}$

Example 3.38. Special Polynomials.

For the limit

${\displaystyle \underset{x\to 8}{\lim }\frac{x^{1∕3}-2}{x-8}\text{,}}$

we use the factorization of the difference of cubes.

${\displaystyle \begin{array}{ll}{\displaystyle \frac{x^{1/3}-2}{x-8}}\hfill & ={\displaystyle \frac{x^{1/3}-2}{{\left(x^{1/3}\right)}^3-2^3}}\hfill \\ \hfill & ={\displaystyle \frac{x^{1/3}-2}{\left(x^{1/3}-2\right)\left(x^{2/3}+2x^{1/3}+2^2\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{x^{2/3}+2x^{1/3}+2^2}}\text{for }x\ne 8\hfill \end{array}}$

Hence, by Proposition 3.16 and Theorem 3.14,

${\displaystyle \underset{x\to 8}{\lim }\frac{x^{1∕3}-2}{x-8}=\underset{x\to 8}{\lim }\frac{1}{x^{2∕3}+2x^{1∕3}+2^2}=\frac{1}{12}\text{.}}$

Tutorial 1: Evaluation of a limit of a discontinuous function at 3, by algebraic manipulation.