Objectives
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Assess when limit rules do not suffice to compute limits.
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Use algebraic techniques to help compute limits.
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Review algebraic techniques of factoring, clearing denominators, and conjugate radical expressions.
Section 1.13 Limits: algebraic technique
It is easy to forget that our limit rules all have restrictive conditions that limit their use: namely, we need the limit of the component functions involved to exist, and in the case of the quotient rule, we need the limit of the denominator function to be nonzero. These restrictions put significant limitations on the extent to which we can apply limit rules to compute limits, as the following simple example illustrates.
Example 1.13.1. Quotient rule does not apply.
Compute the limit below. Your answer should be a chain of equalities with steps justified.
\begin{equation*}
\lim_{x\to 1}\frac{x^2-1}{x-1}
\end{equation*}
Solution.
Note that since the limit of the denominator function is 0, we are not able to use the quotient (or rational function) rule. We begin instead with some algebra:
\begin{align*}
\lim_{x\to 1}\frac{x^2-1}{x-1}\amp =\lim_{x\to 1}\frac{(x-1)(x+1)}{x-1} \\
\amp = \lim_{x\to 1}x+1 \amp \text{(repl. rule)}\\
\amp =2 \amp \text{(poly eval.)}\text{.}
\end{align*}
As we saw in Example 1.13.1, there are limits to the usefulness of our limit rules. In many cases, if we run into a dead end, some algebraic manipulation of the function expression can help us find a way forward. The following examples serve to illustrate this technique and remind you of some common algebraic techniques: e.g., factoring, clearing denominators, and simplifying expressions with radicals.
Example 1.13.2. Algebraic limit technique: factoring.
Compute
\begin{equation*}
\lim_{v\to 2}\frac{v^3-8}{v^4-16}\text{.}
\end{equation*}
Your answer should be a chain of equalities with steps justified.
Solution.
Note that the limit of the denominator expression as \(v\to 2\) is zero, and thus we cannot directly use the quotient rule. Thus we begin with some algebraic manipulation. The key algebraic technique here is factoring.
\begin{align*}
\lim_{v\to 2}\frac{v^3-8}{v^4-16} \amp=\lim_{v\to 2}\frac{(v-2)(v^2+2v+4)}{(v^2-4)(v^2+4)} \\
\amp = \lim_{v\to 2}\frac{(v-2)(v^2+2v+4)}{(v-2)(v+2)(v^2+4)}\\
\amp = \lim_{v\to 2}\frac{v^2+2v+4}{(v+2)(v^2+4)} \amp \text{(repl. rule)}\\
\amp = \frac{2^2+2\cdot 2+4}{(2+2)(2^2+4)} \amp \text{(rational eval.)}\\
\amp = \frac{12}{32}=\frac{3}{8}\text{.}
\end{align*}
Example 1.13.3. Algebraic limit technique: clear denominator.
Compute
\begin{equation*}
\lim\limits_{t\to 1}\frac{t-1}{\frac{t^2}{t+2}-\frac{1}{3}}\text{.}
\end{equation*}
Your answer should be a chain of equalities with steps justified.
Solution.
Again, the quotient rule is not available to us, as the limit of the denominator function is zero. Accordingly, we begin with some algebra.
\begin{align*}
\lim\limits_{t\to 1}\frac{t-1}{\frac{t^2}{t+2}-\frac{1}{3}} \amp =
\lim\limits_{t\to 1}\frac{t-1}{\frac{3t^2-t-2}{3(t+2)}} \amp \text{(like denominators)}\\
\amp =\lim\limits_{t\to 1}\frac{3(t-1)(t+2)}{3t^2-t-2}\\
\amp =\lim\limits_{t\to 1}\frac{3(t-1)(t+2)}{(3t+2)(t-1)} \\
\amp =\lim\limits_{t\to 1}\frac{3(t+2)}{3t+2} \amp \text{(repl. rule)}\\
\amp = \frac{9}{5} \amp \text{(rational eval.)}
\end{align*}
Example 1.13.4. Algebraic limit technique: radicals.
Compute
\begin{equation*}
\lim_{x\to 0}\frac{\sqrt{x^2+100}-10}{x^2}\text{.}
\end{equation*}
Your answer should be a chain of equalities with steps justified.
Solution.
We use the technique of multiplying the numerator and denominator by a “conjugate radical” expression \(\sqrt{x^2+100}+10\text{:}\)
\begin{align*}
\lim_{x\to 0}\frac{\sqrt{x^2+100}-10}{x^2}\amp =\lim_{x\to 0}\frac{\sqrt{x^2+100}-10}{x^2}\frac{\sqrt{x^2+100}+10}{\sqrt{x^2+100}+10} \amp \text{(repl. rule)}\\
\amp = \lim_{x\to 0}\frac{x^2+100-100}{x^2(\sqrt{x^2+100}+10)}\\
\amp = \lim_{x\to 0}\frac{1}{\sqrt{x^2+100}+10}\\
\amp = \frac{1}{\sqrt{\lim\limits_{x\to 0}(x^2+100)}+\lim\limits_{x\to 0}10}
\amp \text{(quot., sum, root)}\\
\amp = \frac{1}{\sqrt{100}+10} \amp \text{(poly. eval.)}\\
\amp = \frac{1}{20}\text{.}
\end{align*}
Remark 1.13.5. Assessing application of limit rules.
The example of this section suggest the following general strategy for computing limits:
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First assess whether the given limit can be computed using limit rules. This usually amounts to checking mentally that the limits of component functions actually exist, and no zeros appear in the denominator.
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If it is clear that we cannot simply apply limit rules, try using algebra in combination with the replacement rule to simplify the limit to a form where limit rules and evaluation formulas can be applied.In this case the particular form of the given function often suggests a particular algebraic technique to try: e.g., factoring, clearing denominators, or simplifying radical expressions.
