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Section 1.12 Limit rules

Subsection Limit rules

We now state some useful limit formulas and rules. These will give us a means of breaking down the limit computation of a complicated function into limits of simpler functions. Technically speaking we must prove the validity of each of these rules and formulas; however this would be a fool’s errand until we have a rigorous definition of the limit to work with! Such a definition will be provided in the near future, though even then we will not concern ourselves overly with proofs; we are more interested in learning how to make valid use of the rules.
Our first theorem gives us two simple formulas for computing the limits of particular types of functions: constant functions, and the identity function.
In contrast to the last theorem, our next theorem does not provide any formulas per se, but rather gives us rules governing how limits interact with various function operations like addition, multiplication, etc..

Remark 1.12.3. Limit rules.

It is useful to think of TheoremΒ 1.12.2 as giving us a bunch of algebraic rules for computing limits of functions defined using addition, subtraction, multiplication, etc. For the most part these rules tell us that we can β€œbring the limit into” various operations: for example, the first two rules tell us that we can bring the limit into sums and differences of functions. Or better, using plain English, they tell us that the limit of a sum (of functions) is the sum of the limits, and that the limit of a difference (of functions) is the difference of the limits.
In fact most of the rules in TheoremΒ 1.12.2 can be nicely summarized in plain English (e.g., β€œthe limit of a product is the product of the limits”, β€œthe limit of an \(n\)-th power is the \(n\)-th power of the limit”), and these summaries are helpful for remembering how the limit interacts with function operations.

Remark 1.12.4. Quotient rule.

Assume as in TheoremΒ 1.12.2 that \(\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x)\) exist and consider the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\text{.}\) Mark well that we can only make use of the quotient rule if \(\lim_{x\to a}g(x)\ne 0\text{,}\) in which case we can conclude that the limit of the quotient is the quotient of the limits.
Note further that in the case where \(\lim_{x\to a}g(x)=0\text{,}\) we cannot automatically conclude that the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\) does not exist; it simply the case that we cannot make use of the quotient rule to evaluate this limit. When this happens, we must look to other means for investigating the limit. See ExampleΒ 1.13.1.

Example 1.12.5. Using limit rules.

Compute the following limits. Your answer should be a chain of equalities with steps justified.
  1. \(\displaystyle \lim\limits_{t\to 6}8(t-5)(t-7)\)
  2. \(\displaystyle \lim\limits_{x\to -2}x^3-4x+8\)
  3. \(\displaystyle \lim\limits_{z\to 4}\sqrt{z^2-10}\)
  4. \(\displaystyle \lim\limits_{y\to 2}\frac{y+2}{y^2+5y+6}\)
Solution.
  1. We have
    \begin{align*} \lim\limits_{t\to 6}8(t-5)(t-7) \amp = 8\lim\limits_{t\to 6}(t-5)(t-7) \amp \text{(scal. mult. rule)} \\ \amp = 8\left(\lim\limits_{t\to 6}(t-5)\right)\left(\lim\limits_{t\to 6}(t-7)\right) \amp \text{(prod. rule)} \\ \amp = 8(6-5)(6-7) \amp \text{(ident. formula)} \\ \amp = 8(1)(-1)=-8\text{.} \end{align*}
  2. We have
    \begin{align*} \lim\limits_{x\to -2}x^3-4x+8 \amp = \lim\limits_{x\to -2}x^3-\lim\limits_{x\to -2}4x+\lim\limits_{x\to -2}8 \amp \text{(sum rule)} \\ \amp = (\lim\limits_{x\to -2}x)^3-4\lim\limits_{x\to -2}x+8 \amp \text{(power, scal. mult., const. formula)} \\ \amp = (-2)^3-4(-2)+8 \amp \text{(ident. formula)} \\ \amp = -8+8+8=8\text{.} \end{align*}
  3. We have
    \begin{align*} \lim\limits_{z\to 4}\sqrt{z^2-10} \amp = \sqrt{\lim\limits_{z\to 4}(z^2-10)} \amp \text{(root rule)} \\ \amp = \sqrt{\lim\limits_{z\to 4}z^2-\lim\limits_{z\to 4}10} \amp \text{(sum rule)} \\ \amp = \sqrt{(\lim\limits_{z\to 4}z)^2-10} \amp \text{(power, const. formula)} \\ \amp = \sqrt{4^2-10} \amp \text{(ident. formula)} \\ \amp = \sqrt{6}\text{.} \end{align*}
  4. We have
    \begin{align*} \lim\limits_{y\to 2}\frac{y+2}{y^2+5y+6} \amp = \frac{\lim\limits_{y\to 2}(y+2)}{\lim\limits_{y\to 2}(y^2+5y+6)} \amp \text{(quot. rule)} \\ \amp = \frac{\lim\limits_{y\to 2}y+\lim\limits_{y\to 2}2}{\lim\limits_{y\to 2}y^2+5\lim\limits_{y\to 2}y+\lim\limits_{y\to 2}6} \amp \text{(sum, scal. mult. rule)} \\ \amp = \frac{2+2}{2^2+5(2)+6} \amp \text{(ident. formula)} \\ \amp = \frac{4}{4+10+6} \amp \text{(ident. formula)} \\ \amp = \frac{4}{20} = \frac{1}{5} \text{.} \end{align*}
You may have noticed in the last example that although we had to apply several limit rules before we could compute \(\lim\limits_{x\to -2}x^3-4x+8\text{,}\) when in the end the value of that limit was simply the result of evaluating the polynomial \(p(x)=x^3-4x+8\) at \(x=-2\text{:}\) i.e., in the end we found that
\begin{equation*} \lim\limits_{x\to -2}p(x)=p(-2)=(-2)^3-4(-2)+8=-8+8+8=8\text{.} \end{equation*}
Using the integer power rule, scalar multiplication rule, and sum rule in sequence, it is easy to see that the same is true for any polynomial \(p(x)=\anpoly\text{.}\) This is the fundamental idea behind the polynomial evaluation formula (1.71) in the theorem below. Throwing the quotient rule into the mix, we can immediately generalize this result to rational functions, since these are just quotients of polynomials. Indeed, give a rational function
\begin{equation*} f(x)=\frac{p(x)}{q(x)}\text{,} \end{equation*}
we have
\begin{align*} \lim\limits_{x\to a}f(x) \amp = \frac{\lim\limits_{x\to a}p(x)}{\lim\limits_{x\to a}q(x)} \amp \text{(quot. rule)} \\ \amp = \frac{p(a)}{q(a)} \amp \text{(poly. eval.)} \text{,} \end{align*}
as long as \(q(a)\ne 0\text{.}\) This shows how the rational function evaluation formula (1.72) follows from the polynomial evaluation formula (1.71). We make these two evaluation formulas official in the form of a theorem.

Example 1.12.7. Polynomial and rational function evaluation.

Give a simplified computation of the limit
\begin{equation*} \lim\limits_{y\to 2}\frac{y+2}{y^2+5y+6} \end{equation*}
Solution.
Since
\begin{equation*} f(y)=\frac{y+2}{y^2+5y+6}=\frac{p(y)}{q(y)} \end{equation*}
is a rational function, and since the denominator polynomial \(q(y)=y^2+5y+6\) satisfies \(q(2)\ne 0\text{,}\) we may apply the rational function evaluation formula:
\begin{align*} \lim\limits_{y\to 2}\frac{p(y)}{q(y)}\amp =\frac{p(2)}{q(2)} \\ \amp=\frac{2+2}{2^2+5\cdot 2+6}\\ \amp =\frac{4}{20}\\ \amp =\frac{1}{5}\text{.} \end{align*}

Warning 1.12.8. Rational function evaluation.

As useful as TheoremΒ 1.12.6 is, mark well the restriction on its application: we can only use this evaluation rule (currently) on polynomials \(p(x)=\anpoly\) and rational functions
\begin{equation*} f(x)=\frac{\anpoly}{\bmpoly}\text{.} \end{equation*}
ExampleΒ 1.12.7 below is a nice illustration of this limitation. As tempted as we might be to compute the limit
\begin{equation*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \end{equation*}
simply by evaluating the function
\begin{equation*} f(x)= \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \end{equation*}
at \(x=-2\text{,}\) we currently have no evaluation rule that allows us to do so directly. We had to use limit rules to simplify the limit expression to a point where we could apply the polynomial evaluation formula.

Example 1.12.9. Using limit rules.

Compute the limit below. Your answer should be a chain of equalities with steps justified.
\begin{equation*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \end{equation*}
Solution.
\begin{align*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \amp = \frac{\lim\limits_{x\to -2}(x^3+x+1)}{\lim\limits_{x\to -2}\sqrt[3]{4x^2+11}} \amp \text{(quot. rule)}\\ \amp =\frac{(-2)^3+(-2)+1}{\sqrt[3]{\lim\limits_{x\to -2}(4x^2+11)}} \amp \text{(poly. eval./root rule)}\\ \amp =\frac{-9}{\sqrt[3]{4(-2)^2+11}} \amp \text{(poly. eval.)}\\ \amp =\frac{-9}{\sqrt[3]{27}}\\ \amp =-\frac{9}{3}=-3\text{.} \end{align*}

Remark 1.12.10. Limit rules and limit formulas.

It is useful to distinguish between limit rules and limit formulas. The various limit rules given in TheoremΒ 1.12.2 gives us a means of reducing a complicated limit computation to a sequence of simpler limit computations. At some point, however, we have to be able to compute these limits!
This is where our limit formulas come into play. Currently the polynomial and rational function evaluation formulas are our most powerful means of computing limits. Notice that these formulas subsume the formulas for constant functions and identity functions given in TheoremΒ 1.12.1, as constant functions and the identity function \(f(x)=x\) are both examples of polynomials (of degree 0 and 1, respectively).
Why do we call these limit formulas evaluation formulas? The formulas for polynomials and rational functions are both of the form
\begin{equation*} \lim\limits_{x\to a}f(x)=f(a)\text{.} \end{equation*}
These formulas instruct us to compute the limit \(\lim\limits_{x\to a}f(x)\) simply by computing \(f(a)\) evaluating the function \(f\) at \(a\text{:}\) that is, by evaluating the function \(f\) at the limit point \(a\text{.}\) Currently we only have evaluation formulas for polynomials and rational functions, but soon we will be able to extend this to a vastly larger class of functions: namely, to all continuous functions. Stay tuned.
We end this section with one more limit rule related to the absolute value function. The proof of this result serves as another example of using limit rules.

Proof.

We will derive this rule by using the identity \(\abs{x}=\sqrt{x^2}\) for all \(x\in \R\text{.}\) We have
\begin{align*} \lim\limits_{x\to a}\abs{f(x)} \amp = \lim\limits_{x\to a}\sqrt{(f(x))^2} \amp (\sqrt{x^2}=\abs{x})\\ \amp = \sqrt{\lim\limits_{x\to a}(f(x))^2} \amp \text{(root rule)} \\ \amp = \sqrt{(\lim\limits_{x\to a}f(x))^2} \amp \text{(power rule)} \\ \amp = \abs{\lim\limits_{x\to a}f(x)} \amp (\sqrt{x^2}=\abs{x}) \text{,} \end{align*}
as claimed.