4.10 The Other Indeterminate Forms

Table of Contents

Review on Limit Laws

Let’s review algebraic operations on limits:

Let $L$ and $M$ be two finite numbers. The following relationships are true as $x \to a$ , $x \to a^{+}$ , $x \to a^{-}$ , $x \to + \infty$ , or $x \to - \infty$

Sum:

Product:

Quotient:

The Indeterminate Forms

Although the theorems tell us a great deal about the behavior of combined functions, there are four cases that the theorems are silent about, specifically the cases in the above tables where there is a question mark in front of. These four cases are denoted by $\infty - \infty, 0 \cdot (\pm \infty), \frac{0}{0}, \frac{\pm \infty}{\pm \infty}$

and are called the indeterminate forms. The value of the indeterminate forms cannot be predicted in advance. Each case may take any value (including $+ \infty$ , $- \infty$ ), or may fail to exist.

Previously, we learned how to evaluate the limits corresponding to the 0/0 form. In this section, we deal with the other three indeterminate forms.

In algebra, 0/0 is not defined and infinity is not a number. We should emphasize that 0/0, $0 \cdot (\pm \infty)$ , and so on are just shorthands for the limits shown in the above tables.

For example, if $f (x) = x^{2} - 1$ and $g (x) = x - 1$ , then $f, g \to 0$ as $x \to 1$ . We cannot divide $f (x)$ by number 0 because division by zero is not defined. However, we can divide $f (x)$ by $g (x)$ because although $g (x)$ approaches zero as $x \to 0$ , $g (x)$ is never exactly zero unless $x = 1$ . So the fraction $f (x) / g (x)$ is defined for $x \neq 1$ , and we can find $lim_{x \to 1} \frac{f (x)}{g (x)}$ because in the limit $x$ gets closer and closer to 1 but never equal 1. Here we say the limit of $f (x) / g (x)$ as $x \to 1$ is the indeterminate form of type 0/0: $\begin{aligned} lim_{x \to 1} \frac{f (x)}{g (x)} & = lim_{x \to 1} \frac{x^{2} - 1}{(x - 1)} \\ = lim_{x \to 1} \frac{(x - 1) (x + 1)}{(x - 1)} \\ = lim_{x \to 1} (x + 1) \\ = 2. \end{aligned}$
We usually drop the $\pm$ signs when we want to refer to the $0 \dot{(} \pm \infty)$ and $\pm \infty / \pm \infty$ indeterminate forms.
There are three more indeterminate forms $0^{0}, \infty^{0}, and 1^{\infty} .$ In total, there are 7 indeterminate forms. These three indeterminate forms will be considered in Chapter 6.

Resolving the Indeterminate Forms

Resolving an indeterminate form means finding the limit.

If $f$ and $g$ are two fractions, find a common denominator, convert them to one indeterminate quotient (often 0/0 or infinity divided by infinity), and then simplify the result.

Example 1

Find $lim_{x \to 1} (\frac{2}{x^{2} - 1} - \frac{1}{x - 1}) .$

Solution

When $x \to 1$ and $x > 1$ (that is, $x \to 1^{+})$ , both $x^{2} - 1$ and $x - 1$ approach zero through positive values. So $lim_{x \to 1^{+}} \frac{2}{x^{2} - 1} \overset{[\frac{2}{0^{+}}]}{=} + \infty, lim_{x \to 1^{+}} \frac{1}{x - 1} \overset{[\frac{1}{0^{+}}]}{=} + \infty$ and we have an indeterminate form $\infty - \infty$ . When $x \to 1$ and $x < 1$ (or $x \to 1^{-})$ , then both $x^{2} - 1$ and $x - 1$ approach zero through negative values. So $lim_{x \to 1^{-}} \frac{2}{x^{2} - 1} \overset{[\frac{2}{0^{-}}]}{=} - \infty, lim_{x \to 1^{+}} \frac{1}{x - 1} \overset{[\frac{1}{0^{-}}]}{=} - \infty$ and we have again an indeterminate form $- \infty - (- \infty)$ or $- \infty + \infty$ .

To evaluate the limit, we find the common denominator. Because $x^{2} - 1 = (x - 1) (x + 1)$ [Recall the identity $A^{2} - B^{2} = (A - B) (A + B)$ ], we have $\begin{aligned} lim_{x \to 1} (\frac{2}{x^{2} - 1} - \frac{1}{x - 1}) & = lim_{x \to 1} (\frac{2}{(x - 1) (x + 1)} - \frac{x + 1}{(x - 1) (x + 1)}) \\ = lim_{x \to 1} \frac{2 - (x + 1)}{(x - 1) (x + 1)} \\ = lim_{x \to 1} \frac{1 - x}{\underset{- 1}{(x - 1)} (x + 1)} \\ = lim_{x \to 1} \frac{- 1}{x + 1} \\ = - \frac{1}{2} . \end{aligned}$ Here we did not have to consider the one-sided limits ( $x \to 1^{+}$ and $x \to 1^{-}$ ) separately.

Example 2

Find $lim_{x \to 0} (\csc x - \cot x)$

Solution

Because $\csc x = \frac{1}{\sin x}, and \cot x = \frac{\cos x}{\sin x}$

and $\sin x \to 0 as x \to 0$ both $\csc x$ and $\cot x$ approach infinity and we deal with $\infty - \infty$ form (You can investigate that $lim_{x \to 0^{+}} \csc x = lim_{x \to 0^{+}} \cot x = + \infty$ and $lim_{x \to 0^{-}} \csc x = lim_{x \to 0^{-}} \cot x = - \infty$ ). So we convert the given expression into one fraction: $\begin{aligned} lim_{x \to 0} (\csc x - \cot x) & = lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x}) \\ = lim_{x \to 0} \frac{1 - \cos x}{\sin x} . \end{aligned}$ Because $1 - \cos x \to 0 and \sin x \to 0 as x \to 0$ now we have to evaluate a limit of the form 0/0. There are several ways to evaluate $lim_{x \to 0} \frac{1 - \cos x}{\sin x}$ . Probably the easiest way is to use L’Hôpital’s rule that we will learn later. Here are some other ways:

Method a: Divide both the denominator and numerator by $x$ and use the following limits that we learned in the section on Theorems for Calculating Limits $lim_{x \to 0} \frac{1 - \cos x}{x} = 0, lim_{x \to 0} \frac{\sin x}{x} = 1$ Therefore $\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{\sin x} & = lim_{x \to 0} \frac{\frac{1 - \cos x}{x}}{\frac{\sin x}{x}} \\ = \frac{lim_{x \to 0} \frac{1 - \cos x}{x}}{lim_{x \to 0} \frac{\sin x}{x}} \\ = \frac{0}{1} = 0. \end{aligned}$

Method b: We can use the half-angle formula (See the section on Trigonomertic Identities): $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ and the double angle formula (See the section on Trigonomertic Identities): $\sin 2 θ = 2 \sin θ \cos θ$ Now let $x = 2 θ$ . Thus $1 - \cos x = 2 \sin^{2} (\frac{x}{2})$ and $\sin x = 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})$

Therefore: $\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{\sin x} & = lim_{x \to 0} \frac{2 \sin^{2} (\frac{x}{2})}{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} \\ = lim_{x \to 0} \frac{\sin (\frac{x}{2})}{\cos (\frac{x}{2})} = \frac{0}{1} = 0. \end{aligned}$

Method c: Multiply both the numerator and denominator by $1 + \cos x$ and use the identities $(A - B) (A + B) = A^{2} - B^{2}$ and $\sin^{2} θ + \cos^{2} θ = 1$ : $\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{\sin x} & = lim_{x \to 0} (\frac{1 - \cos x}{\sin x} \cdot \frac{1 + \cos x}{1 + \cos x}) \\ = lim_{x \to 0} \frac{1 - \cos^{2} x}{\sin x (1 + \cos x)} \\ = lim_{x \to 0} \frac{\sin^{2} x}{\sin x (1 + \cos x)} \\ = lim_{x \to 0} \frac{\sin x}{1 + \cos x} \\ = \frac{\sin 0}{1 + \cos 0} = \frac{0}{2} = 0. \end{aligned}$

Limits of Polynomials and Rational Functions as $x \to \pm \infty$

The end behavior of polynomials is the same as the end behavior of their leading term (= the highest degree term). That is, if $a_{n} \neq 0$ then $lim_{x \to \pm \infty} (a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}) = lim_{x \to \pm \infty} (a_{n} x^{n})$

Read why the above equation is true

To justify these results, let’s factor out the highest power of $x$ and then use the fact that $lim_{x \to \pm \infty} \frac{1}{x^{r}} = 0$ for $r > 0$ : $\begin{aligned} lim_{x \to \pm \infty} (a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}) & = lim_{x \to \pm \infty} (x^{n} (a_{n} + \frac{a_{n - 1}}{x} + \dots + \frac{a_{0}}{x^{n}})) \\ = (lim_{x \to \pm \infty} x^{n}) (lim_{x \to \pm \infty} (a_{n} + \frac{a_{n - 1}}{x} + \dots + \frac{a_{0}}{x^{n}})) \\ = (lim_{x \to \pm \infty} x^{n}) (a_{n} + 0 + \dots + 0 + 0) \\ = a_{n} (lim_{x \to \pm \infty} x^{n}) \\ = lim_{x \to \pm \infty} (a_{n} x^{n}) \end{aligned}$

Example 3

Find

(a) $lim_{x \to - \infty} (- 3 x^{5} + 8 x^{2} + 37)$

(b) $lim_{x \to \infty} (12 + 9 x + 3 x^{3} + 1200 x^{5} - x^{6})$

Solution

(a) $\begin{aligned} lim_{x \to - \infty} (- 3 x^{5} + 8 x^{2} + 37) & = lim_{x \to - \infty} (- 3 x^{5}) = - 3 lim_{x \to - \infty} x^{5} \\ = - 3 (- \infty) \\ = \infty \end{aligned}$ (b) $\begin{aligned} lim_{x \to \infty} (12 + 9 x + 3 x^{3} + 1200 x^{5} - x^{6}) & = lim_{x \to \infty} (- x^{6}) \\ = - lim_{x \to \infty} x^{6} \\ = - (+ \infty) \\ = - \infty \end{aligned}$

With the same reasoning that we used for polynomials, we can say if $a_{n} \neq 0$ and $b_{m} \neq 0$ , then $lim_{x \to \pm \infty} \frac{a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{n - 1} x + a_{0}}{b_{m} x^{m} + b_{m - 1} x^{m - 1} + \dots + b_{m - 1} x + b_{0}} = lim_{x \to \pm \infty} \frac{a_{n} x^{n}}{b_{m} x^{m}}$

and $lim_{x \to \pm \infty} \frac{a_{n} x^{n}}{b_{m} x^{m}} = {\begin{cases} + \infty or - \infty & if n > m \\ \frac{a_{n}}{b_{m}} & if n = m \\ 0 & if n < m \end{cases}$

Show the reasoning

The reasoning is $\begin{aligned} lim_{x \to \pm \infty} \frac{a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}}{b_{m} x^{m} + b_{m - 1} x^{m - 1} + \dots + b_{0}} & = lim_{x \to \pm \infty} \frac{x^{n} (a_{n} + \frac{a_{n - 1}}{x} + \dots + \frac{a_{0}}{x^{n}})}{x^{m} (b_{m} + \frac{b_{m - 1}}{x} + \dots + \frac{b_{0}}{x^{m}})} \\ = \frac{(lim_{x \to \pm \infty} x^{n}) (lim_{x \to \pm \infty} (a_{n} + \frac{a_{n - 1}}{x} + \dots + \frac{a_{0}}{x^{n}}))}{(lim_{x \to \pm \infty} x^{m}) (lim_{x \to \pm \infty} (b_{m} + \frac{b_{m - 1}}{x} + \dots + \frac{b_{0}}{x^{m}}))} \\ = \frac{(lim_{x \to \pm \infty} x^{n}) (a_{n} + 0 + \dots + 0 + 0)}{(lim_{x \to \pm \infty} x^{m}) (b_{m} + 0 + \dots + 0 + 0)} \\ = \frac{lim_{x \to \pm \infty} a_{n} x^{n}}{lim_{x \to \pm \infty} b_{m} x^{m}} \\ = lim_{x \to \pm \infty} \frac{a_{n} x^{n}}{b_{m} x^{m}} . \end{aligned}$

Example 4

Find

(a) $lim_{x \to \infty} \frac{3 + 5 x}{1 - 2 x}$

(b) $lim_{x \to - \infty} \frac{3 x^{3} - 4 x^{2} + x + 3}{- x^{3} + 2 x^{2}}$

(d) $lim_{x \to - \infty} \frac{x^{7} + 1}{- x^{6} + x^{3} + 9}$

Solution

(a) $lim_{x \to \infty} \frac{3 + 5 x}{1 - 2 x} = lim_{x \to \infty} \frac{5 x}{- 2 x} = - \frac{5}{2}$

(b) $\begin{aligned} lim_{x \to - \infty} \frac{3 x^{3} - 4 x^{2} + x + 3}{- x^{3} + 2 x^{2}} & = lim_{x \to - \infty} \frac{3 x^{3}}{- x^{3}} \\ = lim_{x \to - \infty} (- 3) = - 3 \end{aligned}$

(c) $\begin{aligned} lim_{x \to \infty} \frac{15 + 4 x - 5 x^{3} + 3 x^{4}}{1 + 16 x^{2} - x^{5}} & = lim_{x \to \infty} \frac{3 x^{4}}{- x^{5}} \\ = lim_{x \to \infty} \frac{- 3}{x} \\ = - 3 lim_{x \to \infty} \frac{1}{x} \\ = - 3 (0) \\ = 0 \end{aligned}$

(d) $\begin{aligned} lim_{x \to - \infty} \frac{x^{7} + 1}{- x^{6} + x^{3} + 9} & = lim_{x \to - \infty} \frac{x^{7}}{- x^{6}} \\ = lim_{x \to - \infty} (- x) \\ = - lim_{x \to - \infty} x \\ = - (- \infty) \\ = \infty \end{aligned}$

Example 5

Find $lim_{x \to + \infty} (\frac{2 x^{3}}{x^{2} - 4} - \frac{2 x^{2}}{x - 1})$

Solution

Notice that $lim_{x \to + \infty} \frac{2 x^{3}}{x^{2} - 4} = lim_{x \to + \infty} \frac{2 x^{3}}{x^{2}} = lim_{x \to + \infty} 2 x = + \infty$ and $lim_{x \to + \infty} \frac{2 x^{2}}{x - 1} = lim_{x \to + \infty} \frac{2 x^{2}}{x} = lim_{x \to + \infty} 2 x = + \infty .$

So we cannot use the Difference Rule and say $lim_{x \to \infty} (\frac{2 x^{3}}{x^{2} - 4} - \frac{2 x^{2}}{x - 1}) = lim_{x \to \infty} \frac{2 x^{3}}{x^{2} - 4} - lim_{x \to \infty} \frac{2 x^{2}}{x - 1} = \infty - \infty$ because $\infty - \infty$ is an indeterminate form. We have to subtract the fractions and express the result as a rational function: $\frac{2 x^{3}}{x^{2} - 4} - \frac{2 x^{2}}{x - 1} = \frac{2 x^{3} (x - 1)}{(x^{2} - 4) (x - 1)} - \frac{2 x^{2} (x^{2} - 4)}{(x - 1) (x^{2} - 4)}$ Thus $\begin{aligned} \frac{2 x^{3}}{x^{2} - 4} - \frac{x^{2}}{2 x - 1} & = \frac{2 x^{4} - 2 x^{3} - 2 x^{4} + 8 x^{2}}{x^{3} - x^{2} - 4 x + 4} \\ = \frac{- 2 x^{3} + 8 x^{2}}{x^{3} - x^{2} - 4 x + 4} \end{aligned}$ and $\begin{aligned} lim_{x \to \infty} (\frac{2 x^{3}}{x^{2} - 4} - \frac{2 x^{2}}{x - 1}) & = lim_{x \to \infty} \frac{- 2 x^{3} + 8 x^{2}}{x^{3} - x^{2} - 4 x + 4} \\ = lim_{x \to \infty} \frac{- 2 x^{3}}{x^{3}} \\ = - 2. \end{aligned}$

Limits Involving Radicals

Example 6

Find $lim_{x \to \infty} \frac{x^{2} - 4 \sqrt{x} + 3 x}{- 3 x^{2} + 5 x^{2 / 3} + 1}$

Solution

Let’s factor out the highest power of $x$ occurring in the denominator and that in the numerator $\begin{aligned} lim_{x \to \infty} \frac{x^{2} - 4 \sqrt{x} + 3 x}{- 3 x^{2} + 5 x^{2 / 3} + 1} & = lim_{x \to \infty} \frac{x^{2} (1 - 4 x^{1 / 2 - 2} + 3 x^{- 1})}{x^{2} (- 3 + 5 x^{2 / 3 - 2} + x^{- 2})} \\ = lim_{x \to \infty} \frac{1 - \frac{4}{x^{3 / 2}} + \frac{3}{x}}{- 3 + \frac{5}{x^{4 / 3}} + \frac{1}{x^{2}}} \\ = \frac{lim_{x \to \infty} (1 - \frac{4}{x^{3 / 2}} + \frac{3}{x})}{lim_{x \to \infty} (- 3 + \frac{5}{x^{4 / 3}} + \frac{1}{x^{2}})} \\ = \frac{lim_{x \to \infty} 1 - 4 lim_{x \to \infty} \frac{1}{x^{3 / 2}} + 3 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} (- 3) + 5 lim_{x \to \infty} \frac{1}{x^{4 / 3}} + lim_{x \to \infty} \frac{1}{x^{2}}} \end{aligned}$ Now recall part 1 of Theorem [[thm:Ch4-Important-Limits-at-Infinity]] that tells us that $lim_{x \to \pm \infty} \frac{1}{x^{r}} = 0$ ( $r > 0$ is a rational number). Thus, $\begin{aligned} \frac{lim_{x \to \infty} 1 - 4 lim_{x \to \infty} \frac{1}{x^{3 / 2}} + 3 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} (- 3) + 5 lim_{x \to \infty} \frac{1}{x^{4 / 3}} + lim_{x \to \infty} \frac{1}{x^{2}}} & = \frac{1 - 4 (0) + 3 (0)}{- 3 + 5 (0) + 0} \\ = - \frac{1}{3} . \end{aligned}$

Example 7

Find $lim_{x \to \infty} \frac{2 \sqrt{x} - 3 \sqrt[3]{x} + 4 \sqrt[5]{x}}{\sqrt{3 x + 1} - \sqrt[3]{2 x + 7}}$

Solution

Let’s factor out the highest power of $x$ , which is here $\sqrt{x} = x^{1 / 2}$ : $\begin{aligned} lim_{x \to \infty} \frac{2 \sqrt{x} - 3 \sqrt[3]{x} + 4 \sqrt[5]{x}}{\sqrt{3 x + 1} - \sqrt[3]{2 x + 7}} & = lim_{x \to \infty} \frac{\sqrt{x} (2 - x^{1 / 3 - 1 / 2} + 4 x^{1 / 5 - 1 / 2})}{\sqrt{x} (\frac{\sqrt{3 x + 1}}{\sqrt{x}} - \frac{\sqrt[3]{2 x + 7}}{\sqrt{x}})} \\ = lim_{x \to \infty} \frac{\sqrt{x} (2 - x^{- 1 / 6} + 4 x^{- 3 / 10})}{\sqrt{x} (\sqrt{\frac{3 x + 1}{x}} - \sqrt[3]{\frac{2 x + 7}{x^{3 / 2}}})} \\ = lim_{x \to \infty} \frac{2 - \frac{1}{x^{1 / 6}} + \frac{4}{x^{3 / 10}}}{\sqrt{3 + \frac{1}{x}} - \sqrt[3]{\frac{2}{x^{1 / 2}} + \frac{7}{x^{3 / 2}}}} \\ = \frac{lim_{x \to \infty} 2 - lim_{x \to \infty} \frac{1}{x^{1 / 6}} + 4 lim_{x \to \infty} \frac{1}{x^{3 / 10}}}{\sqrt{lim_{x \to \infty} 3 + lim_{x \to \infty} \frac{1}{x}} - \sqrt[3]{2 lim_{x \to \infty} \frac{1}{x^{1 / 2}} + 7 lim_{x \to \infty} \frac{1}{x^{3 / 2}}}} \end{aligned}$ Using the facts that $lim_{x \to \infty} c = c$ (where $c$ is a constant), $lim_{x \to \infty} \sqrt[n]{f (x)} = \sqrt[n]{lim_{x \to \infty} f (x)},$ and $lim_{x \to \infty} \frac{1}{x^{r}} = 0$ (where $r > 0$ is a rational number), we get $\begin{aligned} \frac{lim_{x \to \infty} 2 - lim_{x \to \infty} \frac{1}{x^{1 / 6}} + 4 lim_{x \to \infty} \frac{1}{x^{3 / 10}}}{\sqrt{lim_{x \to \infty} 3 + lim_{x \to \infty} \frac{1}{x}} - \sqrt[3]{2 lim_{x \to \infty} \frac{1}{x^{1 / 2}} + 7 lim_{x \to \infty} \frac{1}{x^{3 / 2}}}} & = \frac{2 - 0 + 4 (0)}{\sqrt{3 + 0} - \sqrt[3]{2 (0) + 7 (0)}} \\ = \frac{2}{\sqrt{3}} . \end{aligned}$

Example 8

Find $lim_{x \to - \infty} \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}}$

Solution

Let’s factor out the highest power of $x$ in the expression under the radical and the highest power of $x$ in the denominator. $\begin{aligned} lim_{x \to - \infty} \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}} & = lim_{x \to - \infty} \frac{\sqrt{x^{4} (4 + \frac{3}{x^{2}} + \frac{1}{x})}}{x^{2} (\frac{1}{x^{2}} - 2)} \\ = lim_{x \to - \infty} \frac{\sqrt{x^{4}} \sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{x^{2} (\frac{1}{x^{2}} - 2)} \end{aligned}$ Because $\sqrt[n]{x^{m}} = x^{m / n}$ $\begin{aligned} lim_{x \to - \infty} \frac{\sqrt{x^{4}} \sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{x^{2} (\frac{1}{x^{2}} - 2)} & = lim_{x \to - \infty} \frac{x^{2} \sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{x^{2} (\frac{1}{x^{2}} - 2)} \\ = lim_{x \to - \infty} \frac{\sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{(\frac{1}{x^{2}} - 2)} \\ = \frac{lim_{x \to - \infty} \sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{lim_{x \to - \infty} (\frac{1}{x^{2}} - 2)} \end{aligned}$

Recall that $lim_{x \to \pm \infty} \sqrt[n]{f (x)} = \sqrt[n]{lim_{x \to \pm \infty} f (x)}$ . Thus $\begin{aligned} \frac{lim_{x \to - \infty} \sqrt{4 + \frac{3}{x^{2}} + \frac{1}{x}}}{lim_{x \to - \infty} (\frac{1}{x^{2}} - 2)} & = \frac{\sqrt{lim_{x \to - \infty} (4 + \frac{3}{x^{2}} + \frac{1}{x})}}{lim_{x \to - \infty} (\frac{1}{x^{2}} - 2)} \\ = \frac{\sqrt{4 + 3 (0) + 0}}{0 - 2} \\ = \frac{2}{- 2} \\ = - 1. \end{aligned}$

The following figure illustrates the graph of $f (x) = \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}}$ .

Figure: Graph of $f (x) = \frac{\sqrt{4 x^{4} + 3 x^{2} + 1}}{1 - 2 x^{2}}$ . It is clear from this graph that $lim_{x \to \pm \infty} f (x) = - 1$ .

Example 9

Find (a) $lim_{t \to \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9}$

(b) $lim_{t \to - \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9}$

Solution

(a) Again we factor out the highest power of the variable $t$ inside the radical and the highest power of $t$ in the denominator $\begin{aligned} lim_{t \to \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9} & = lim_{t \to \infty} \frac{\sqrt{t^{2} (1 + \frac{6}{t^{2}})}}{t (3 - \frac{9}{t})} \\ = lim_{t \to \infty} \frac{\sqrt{t^{2}} \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} \end{aligned}$ Recall that $\sqrt{t^{2}} = | t |$ . So $lim_{t \to \infty} \frac{\sqrt{t^{2}} \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} = lim_{t \to \infty} \frac{| t | \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})}$ Because when $t \to \infty$ , the values under consideration are positive $t > 0$ , $\begin{aligned} lim_{t \to \infty} \frac{| t | \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} & = lim_{t \to \infty} \frac{t \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} \\ = \frac{lim_{t \to \infty} \sqrt{1 + \frac{6}{t^{2}}}}{lim_{t \to \infty} (3 - \frac{9}{t})} \\ = \frac{\sqrt{lim_{t \to \infty} (3 + \frac{6}{t^{2}})}}{lim_{t \to \infty} 3 - lim_{t \to \infty} \frac{9}{t}} \\ = \frac{\sqrt{lim_{t \to \infty} 1 + lim_{t \to \infty} \frac{6}{t}}}{lim_{t \to \infty} 3 - lim_{t \to \infty} \frac{9}{t}} \\ = \frac{\sqrt{1 + 0}}{3 - 0} \\ = \frac{1}{3} . \end{aligned}$ Here we have used the fact that $\frac{1}{t^{r}} \to 0$ as $t \to \pm \infty$ where $r > 0$ is a rational number.

(b) For part b, we follow exactly the same steps that we took in part (a), except when $t \to - \infty$ , the values under consideration are negative $t < 0$ and thus $| t | = - t$ : $\begin{aligned} lim_{t \to - \infty} \frac{\sqrt{t^{2} + 6}}{3 t - 9} & = lim_{t \to - \infty} \frac{\sqrt{t^{2}} \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} \\ = lim_{t \to - \infty} \frac{| t | \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} \\ = lim_{t \to - \infty} \frac{- t \sqrt{1 + \frac{6}{t^{2}}}}{t (3 - \frac{9}{t})} \\ = \frac{- \sqrt{lim_{t \to \infty} 1 + lim_{t \to \infty} \frac{6}{t^{2}}}}{lim_{t \to - \infty} 3 - lim_{t \to \infty} \frac{9}{t}} \\ = \frac{- \sqrt{1 + 0}}{3 - 0} \\ = - \frac{1}{3} . \end{aligned}$

The graph of $f (t) = \frac{\sqrt{t^{2} + 6}}{3 t - 9}$ is shown below.

Figure: Graph of $f (t) = \frac{\sqrt{t^{2} + 6}}{3 t - 9}$ . It is clear from this graph that $f (t) \to \frac{1}{3}$ as $t \to \infty$ and $f (t) \to - 1 / 3$ as $t \to - \infty$ .

Example 10

Find $lim_{x \to - \infty} (\sqrt{x^{4} + 5} - 2 x^{2})$

Solution

Similar to the previous examples, we factor out the highest power of $x$ inside the radical: $lim_{x \to - \infty} (\sqrt{x^{4} + 5} - 2 x^{2}) = lim_{x \to - \infty} (\sqrt{x^{4} (1 + \frac{5}{x^{4}})} - 2 x^{2})$ Because $x^{4} \geq 0$ and $1 + \frac{5}{x^{4}} \geq 0$ , thus $\sqrt{x^{4} (1 + \frac{5}{x^{4}})} = \sqrt{x^{4}} \sqrt{1 + \frac{5}{x^{4}}} = x^{2} \sqrt{1 + \frac{5}{x^{4}}}$ and $\begin{aligned} lim_{x \to - \infty} (\sqrt{x^{4} (1 + \frac{5}{x^{4}})} - 2 x^{2}) & = lim_{x \to - \infty} (x^{2} \sqrt{1 + \frac{5}{x^{4}}} - 2 x^{2}) \\ = lim_{x \to - \infty} (x^{2} (\sqrt{1 + \frac{5}{x^{4}}} - 2)) \\ = (lim_{x \to - \infty} x^{2}) lim_{x \to - \infty} (\sqrt{1 + \frac{5}{x^{4}}} - 2) \\ = (lim_{x \to - \infty} x^{2}) (\sqrt{lim_{x \to - \infty} 1 + lim_{x \to - \infty} \frac{5}{x^{4}}} - lim_{x \to - \infty} 2) \\ = (+ \infty) (\sqrt{1 + 0} - 2) \\ = - \infty \end{aligned}$ The graph of $f (x) = \sqrt{x^{4} + 5} - 2 x^{2}$ is shown below.

Figure: Graph of $f (x) = \sqrt{x^{4} + 5} - 2 x^{2}$ . It is clear from this graph that $f (x) \to - \infty$ as $x \to \pm \infty$ .

Example 11

Find $lim_{x \to - \infty} (\sqrt{x^{4} + 5} - x^{2})$

Solution

The function $f (x) = \sqrt{x^{4} + 5} - x^{2}$ looks very similar to the function in the previous example, so we might be tempted to follow the same procedure. Let’s try that out: $lim_{x \to - \infty} (\sqrt{x^{4} + 5} - x^{2}) = lim_{x \to - \infty} (x^{2} (\sqrt{1 + \frac{5}{x^{4}}} - 1))$ However, because $lim_{x \to - \infty} x^{2} = \infty, lim_{x \to - \infty} (\sqrt{1 + \frac{5}{x^{4}}} - 1) = 0$ we will get the indeterminate limit $0 \cdot \infty$ . Fail!

Instead let’s multiply and divide this function by the conjugate of $\sqrt{x^{4} + 5} - x^{2}$ , which is $\sqrt{x^{4} + 5} + x^{2}$ : $lim_{x \to - \infty} (\sqrt{x^{4} + 5} - x^{2}) = lim_{x \to - \infty} ((\sqrt{x^{4} + 5} - x^{2}) \frac{\sqrt{x^{4} + 5} + x^{2}}{\sqrt{x^{4} + 5} + x^{2}})$ Recall that $(A - B) (A + B) = A^{2} - B^{2}$ . Therefore,

$\begin{aligned} lim_{x \to - \infty} (\frac{(\sqrt{x^{4} + 5} - x^{2}) (\sqrt{x^{4} + 5} + x^{2})}{\sqrt{x^{4} + 5} + x^{2}}) & = lim_{x \to - \infty} \frac{x^{4} + 5 - (x^{2})^{2}}{\sqrt{x^{4} + 5} + x^{2}} \\ = lim_{x \to - \infty} \frac{5}{\sqrt{x^{4} + 5} + x^{2}} \\ = lim_{x \to - \infty} \frac{5}{x^{2} (\sqrt{1 + \frac{5}{x^{4}}} + 1)} \\ = \frac{lim_{x \to - \infty} 5}{lim_{x \to - \infty} x^{2} (\sqrt{lim_{x \to - \infty} 1 + lim_{x \to - \infty} \frac{5}{x^{4}}} + lim_{x \to - \infty} 1)} \\ = \frac{5}{+ \infty (\sqrt{1 + 0} + 1)} = \frac{5}{2 \infty} = \frac{5}{\infty} \\ = 0. \end{aligned}$

The graph of $f (x) = \sqrt{x^{4} + 5} - x^{2}$ is shown below.

Figure: Graph of $f (x) = \sqrt{x^{4} + 5} - x^{2}$ . It is clear from this graph that $f (x) \to 0$ as $x \to \pm \infty$ .

Example 12

Find $lim_{x \to - \infty} (\sqrt{x^{4} + 5 x^{2}} - x^{2})$

Solution

Again factoring out $x^{4}$ will give the indeterminate limit $0 \cdot \infty$ . Let’s try it out $\begin{aligned} lim_{x \to - \infty} (\sqrt{x^{4} + 5 x^{2}} - x^{2}) & = lim_{x \to - \infty} (x^{2} \sqrt{1 + \frac{5}{x^{2}}} - x^{2}) \\ = lim_{x \to - \infty} (x^{2} (\sqrt{1 + \frac{5}{x^{2}}} - 1)) \end{aligned}$ and $lim_{x \to - \infty} x^{2} = \infty$ and $lim_{x \to - \infty} (\sqrt{1 + \frac{5}{x^{2}}} - 1) = 0$ . So like the previous example, let’s multiply and divide by the conjugate of the given function. $\begin{aligned} lim_{x \to - \infty} (\sqrt{x^{4} + 5 x^{2}} - x^{2}) & = lim_{x \to - \infty} ((\sqrt{x^{4} + 5 x^{2}} - x^{2}) \frac{\sqrt{x^{4} + 5 x^{2}} + x^{2}}{\sqrt{x^{4} + 5 x^{2}} + x^{2}}) \\ = lim_{x \to - \infty} \frac{x^{4} + 5 x^{2} - x^{4}}{\sqrt{x^{4} + 5 x^{2}} + x^{2}} \\ = lim_{x \to - \infty} \frac{5 x^{2}}{x^{2} (\sqrt{1 + \frac{5}{x^{2}}} + 1)} \\ = \frac{lim_{x \to - \infty} 5}{\sqrt{lim_{x \to - \infty} 1 + lim_{x \to - \infty} \frac{5}{x^{2}}} + lim_{x \to - \infty} 1} \\ = \frac{5}{\sqrt{1 + 0} + 1} \\ = \frac{5}{2} . \end{aligned}$

The graph of $f (x) = \sqrt{x^{4} + 5 x^{2}} - x^{2}$ is shown below.

Figure: Graph of $f (x) = \sqrt{x^{4} + 5 x^{2}} - x^{2}$ . It is clear from this graph that $f (x) \to 2.5$ as $x \to \pm \infty$ .

Example 13

Find $lim_{x \to - \infty} (\sqrt{2 x^{2} - 1} - 7 x) .$

Solution

Notice that $\begin{aligned} lim_{x \to - \infty} \sqrt{2 x^{2} - 1} & = lim_{x \to - \infty} \sqrt{x^{2} (2 - \frac{1}{x^{2}})} \\ = lim_{x \to - \infty} (| x | \sqrt{2 - \frac{1}{x^{2}}}) & (Recall \sqrt{x^{2}} = | x |) \end{aligned}$ As $x \to - \infty$ , the values under consideration are negative $x < 0$ , thus $| x | = - x$ and $lim_{x \to - \infty} (| x | \sqrt{2 - \frac{1}{x^{2}}}) = lim_{x \to - \infty} (- x \sqrt{2 - \frac{1}{x^{2}}}) = - (- \infty) \sqrt{2 - 0} = + \infty$ and $lim_{x \to - \infty} - 7 x = - 7 lim_{x \to - \infty} x = - 7 (- \infty) = \infty$ Because the sum of two positive infinitely large quantities is also an infinitely large quantity, then $lim_{x \to - \infty} (\sqrt{2 x^{2} - 1} - 7 x) = lim_{x \to - \infty} (\sqrt{2 x^{2} - 1} + (- 7 x)) = \infty .$ In other words, $+ \infty + (+ \infty) = + \infty .$ This is not an indeterminate form.

Review on Limit Laws

The Indeterminate Forms

Resolving the Indeterminate Forms