## 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 $\bbox[#F2F2F2,5px,border:2px solid black]{\infty-\infty,\quad0\cdot(\pm\infty),\quad\frac{0}{0},\quad\frac{\pm\infty}{\pm\infty }}$

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 $$f,g\to0$$ as $$x\to1$$. 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 $$x\to0$$$$g(x)$$ is never exactly zero unless $$x=1$$. So the fraction $$f(x)/g(x)$$ is defined for $$x\neq1$$, and we can find $$\lim_{x\to1}\frac{f(x)}{g(x)}$$ 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 $$x\to1$$ is the indeterminate form of type 0/0: \begin{aligned}\lim_{x\to1}\frac{f(x)}{g(x)} & =\lim_{x\to1}\frac{x^{2}-1}{(x-1)}\\ & =\lim_{x\to1}\frac{(x-1)(x+1)}{(x-1)}\\ & =\lim_{x\to1}(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 $0^{0},\quad\infty^{0},\ \text{and }\ 1^{\infty}.$ 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 $\lim_{x\to1}\left(\frac{2}{x^{2}-1}-\frac{1}{x-1}\right).$

Solution

When $$x\to1$$ and $$x>1$$ (that is, $$x\to1^{+})$$, both $$x^{2}-1$$ and $$x-1$$ approach zero through positive values. So $\lim_{x\to1^{+}}\frac{2}{x^{2}-1}\stackrel{\left[\frac{2}{0^{+}}\right]}{=}+\infty,\quad\lim_{x\to1^{+}}\frac{1}{x-1}\stackrel{\left[\frac{1}{0^{+}}\right]}{=}+\infty$ and we have an indeterminate form $$\infty-\infty$$. When $$x\to1$$ and $$x<1$$ (or $$x\to1^{-})$$, then both $$x^{2}-1$$ and $$x-1$$ approach zero through negative values. So $\lim_{x\to1^{-}}\frac{2}{x^{2}-1}\stackrel{\left[\frac{2}{0^{-}}\right]}{=}-\infty,\quad\lim_{x\to1^{+}}\frac{1}{x-1}\stackrel{\left[\frac{1}{0^{-}}\right]}{=}-\infty$ 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\to1}\left(\frac{2}{x^{2}-1}-\frac{1}{x-1}\right) & =\lim_{x\to1}\left(\frac{2}{(x-1)(x+1)}-\frac{x+1}{(x-1)(x+1)}\right)\\ & =\lim_{x\to1}\frac{2-(x+1)}{(x-1)(x+1)}\\ & =\lim_{x\to1}\frac{\cancel{1-x}}{\underset{-1}{\cancel{(x-1)}}(x+1)}\\ & =\lim_{x\to1}\frac{-1}{x+1}\\ & =-\frac{1}{2}.\end{aligned} Here we did not have to consider the one-sided limits ($$x\to1^{+}$$ and $$x\to1^{-}$$) separately.

Example 2

Find $\lim_{x\to0}(\csc x-\cot x)$

Solution

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

and $\sin x\to 0\ \text{ as }\ x\to 0$ both $$\csc x$$ and $$\cot x$$ approach infinity and we deal with $$\infty-\infty$$ form (You can investigate that $$\lim_{x\to0^{+}}\csc x=\lim_{x\to0^{+}}\cot x=+\infty$$ and $$\lim_{x\to0^{-}}\csc x=\lim_{x\to0^{-}}\cot x=-\infty$$). So we convert the given expression into one fraction: \begin{aligned} \lim_{x\to0}\left(\csc x-\cot x\right) & =\lim_{x\to0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)\\ & =\lim_{x\to0}\frac{1-\cos x}{\sin x}.\end{aligned} Because $1-\cos x\to 0\quad\text{and}\quad \sin x\to 0\quad \text{as }x\to 0$ now we have to evaluate a limit of the form 0/0. There are several ways to evaluate $$\lim_{x\to0}\frac{1-\cos x}{\sin x}$$. 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  $\lim_{x\to0}\frac{1-\cos x}{x}=0,\quad\lim_{x\to0}\frac{\sin x}{x}=1$ Therefore \begin{aligned} \lim_{x\to0}\frac{1-\cos x}{\sin x} & =\lim_{x\to0}\frac{\dfrac{1-\cos x}{x}}{\dfrac{\sin x}{x}}\\ & =\frac{\lim_{x\to0}\dfrac{1-\cos x}{x}}{\lim_{x\to0}\dfrac{\sin x}{x}}\\ & =\frac{0}{1}=0.\end{aligned}

Method b: We can use the half-angle formula (See the section on Trigonomertic Identities): $\sin^{2}\theta=\frac{1-\cos2\theta}{2}$ and the double angle formula (See the section on Trigonomertic Identities): $\sin2\theta=2\sin\theta\cos\theta$ Now let $$x=2\theta$$. Thus $1-\cos x=2\sin^{2}\left(\frac{x}{2}\right)$ and $\sin x=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$

Therefore: \begin{aligned} \lim_{x\to0}\frac{1-\cos x}{\sin x} & =\lim_{x\to0}\frac{2\sin^{2}\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}\\ & =\lim_{x\to0}\frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)}=\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 $$\sin^{2}\theta+\cos^{2}\theta=1$$: \begin{aligned} \lim_{x\to0}\frac{1-\cos x}{\sin x} & =\lim_{x\to0}\left(\frac{1-\cos x}{\sin x}\cdot\frac{1+\cos x}{1+\cos x}\right)\\ & =\lim_{x\to0}\frac{1-\cos^{2}x}{\sin x\ (1+\cos x)}\\ & =\lim_{x\to0}\frac{\sin^{2}x}{\sin x\ (1+\cos x)}\\ & =\lim_{x\to0}\frac{\sin x}{1+\cos x}\\ & =\frac{\sin0}{1+\cos0}=\frac{0}{2}=0.\end{aligned}

### Limits of Polynomials and Rational Functions as $$x\to+\infty$$ or$$x\to-\infty$$

The end behavior of polynomials is the same as the end behavior of their leading term (= the highest degree term). That is, if $$a_{n}\neq0$$ then $\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to\pm\infty}(a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0})=\lim_{x\to\pm\infty}(a_{n}x^{n})}$

#### 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}+\cdots+a_{0}) & =\lim_{x\to\pm\infty}\left(x^{n}\left(a_{n}+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)\right)\\ & =\left(\lim_{x\to\pm\infty}x^{n}\right)\left(\lim_{x\to\pm\infty}\left(a_{n}+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)\right)\\ & =\left(\lim_{x\to\pm\infty}x^{n}\right)\left(a_{n}+0+\cdots+0+0\right)\\ & =a_{n}\left(\lim_{x\to\pm\infty}x^{n}\right)\\ & =\lim_{x\to\pm\infty}(a_{n}x^{n}) \end{aligned}

Example 3

Find

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

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

Solution

(a) \begin{aligned} \lim_{x\to-\infty}(-3x^{5}+8x^{2}+37) & =\lim_{x\to-\infty}(-3x^{5})=-3\lim_{x\to-\infty}x^{5}\\ & =-3(-\infty)\\ & =\infty\end{aligned} (b) \begin{aligned} \lim_{x\to\infty}(12+9x+3x^{3}+1200x^{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 $$a_{n}\neq0$$ and $$b_{m}\neq0$$, then $\bbox[#F2F2F2,5px,border:2px solid black]{\lim_{x\to\pm\infty}\frac{a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{n-1}x+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{m-1}x+b_{0}}=\lim_{x\to\pm\infty}\frac{a_{n}x^{n}}{b_{m}x^{m}}}$

and $\lim_{x\to\pm\infty}\frac{a_{n}x^{n}}{b_{m}x^{m}}=\begin{cases} +\infty\text{ or }-\infty & \text{if }n>m\\[6pt] \dfrac{a_{n}}{b_{m}} & \text{if }n=m\\[6pt] 0 & \text{if }n<m \end{cases}$

#### Show the reasoning

The reasoning is \begin{aligned} \lim_{x\to\pm\infty}\frac{a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{0}} & =\lim_{x\to\pm\infty}\frac{x^{n}\left(a_{n}+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)}{x^{m}\left(b_{m}+\frac{b_{m-1}}{x}+\cdots+\frac{b_{0}}{x^{m}}\right)}\\[6pt] & =\frac{\left({\displaystyle \lim_{x\to\pm\infty}}x^{n}\right)\left({\displaystyle \lim_{x\to\pm\infty}}\left(a_{n}+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)\right)}{\left({\displaystyle \lim_{x\to\pm\infty}}x^{m}\right)\left({\displaystyle \lim_{x\to\pm\infty}}\left(b_{m}+\frac{b_{m-1}}{x}+\cdots+\frac{b_{0}}{x^{m}}\right)\right)}\\[6pt] & =\frac{\left(\lim_{x\to\pm\infty}x^{n}\right)\left(a_{n}+0+\cdots+0+0\right)}{\left(\lim_{x\to\pm\infty}x^{m}\right)\left(b_{m}+0+\cdots+0+0\right)}\\[6pt] & =\frac{\displaystyle{\lim_{x\to\pm\infty}}a_{n}x^{n}}{{\displaystyle \lim_{x\to\pm\infty}}b_{m}x^{m}}\\[6pt] & =\lim_{x\to\pm\infty}\frac{a_{n}x^{n}}{b_{m}x^{m}}.\end{aligned}

Example 4

Find

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

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

(c)  $${\displaystyle \lim_{x\to\infty}\frac{15+4x-5x^{3}+3x^{4}}{1+16x^{2}-x^{5}}}$$

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

Solution

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

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

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

$\lim_{x\to+\infty}\frac{3x-4}{5x-\sin x^{2}}.$

Solution

We notice that $\sin x^{2}$ indefinitely wiggles between $-1$ and $+1$. As $x\to+\infty$, $+1$ or $-1$ is negligible compared to $5x$. So we may say
$\lim_{x\to+\infty}\frac{3x-4}{5x-\sin x^{2}}=\lim_{x\to+\infty}\frac{3x-4}{5x}=\frac{3}{5}.$ Also, we can say when $x$ is large:
$\frac{5x-4}{3x+1}\leq\frac{5x-4}{3x-\sin x^{2}}\leq\frac{5x-4}{3x-1}.$ Because
$\lim_{x\to+\infty}\frac{5x-4}{3x-1}=\frac{5}{3},\qquad\text{and}\qquad\lim_{x\to+\infty}\frac{5x-4}{3x+1}=\frac{5}{3},$ it follows from the Sandwich Theorem that
$\lim_{x\to+\infty}\frac{5x-4}{3x-\sin x^{2}}=\frac{5}{3}.$ The following figure shows the graphs of ${\displaystyle y=\frac{5x-4}{3x+1}}$, ${\displaystyle y=\frac{5x-4}{3x-\sin x^{2}}}$, and ${\displaystyle y=\frac{5x-4}{3x-1}}$.

Example 6

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

Solution

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

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

Example 7

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

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}+3x}{-3x^{2}+5x^{2/3}+1} & =\lim_{x\to\infty}\frac{\cancel{x^{2}}(1-4x^{1/2-2}+3x^{-1})}{\cancel{x^{2}}(-3+5x^{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}\left(1-\frac{4}{x^{3/2}}+\frac{3}{x}\right)}{\lim_{x\to\infty}\left(-3+\frac{5}{x^{4/3}}+\frac{1}{x^{2}}\right)}\\ & =\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 this theorem  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 8

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

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{3x+1}-\sqrt[3]{2x+7}} & =\lim_{x\to\infty}\frac{\sqrt{x}\left(2-x^{1/3-1/2}+4x^{1/5-1/2}\right)}{\sqrt{x}\left(\frac{\sqrt{3x+1}}{\sqrt{x}}-\frac{\sqrt[3]{2x+7}}{\sqrt{x}}\right)}\\[6pt] & =\lim_{x\to\infty}\frac{\cancel{\sqrt{x}}\left(2-x^{-1/6}+4x^{-3/10}\right)}{\cancel{\sqrt{x}}\left(\sqrt{\frac{3x+1}{x}}-\sqrt[3]{\frac{2x+7}{x^{3/2}}}\right)}\\[6pt] & =\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}}}}\\[6pt] & =\frac{\displaystyle{\lim_{x\to\infty}}2-\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{1/6}}+4\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{3/10}}}{\sqrt{\displaystyle{\lim_{x\to\infty}}3+\displaystyle{\lim_{x\to\infty}}\frac{1}{x}}-\sqrt[3]{2\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{1/2}}+7\displaystyle{\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{\displaystyle{\lim_{x\to\infty}}2-\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{1/6}}+4\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{3/10}}}{\sqrt{\displaystyle{\lim_{x\to\infty}}3+\displaystyle{\lim_{x\to\infty}}\frac{1}{x}}-\sqrt[3]{2\displaystyle{\lim_{x\to\infty}}\frac{1}{x^{1/2}}+7\displaystyle{\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 9

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

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{4x^{4}+3x^{2}+1}}{1-2x^{2}} & =\lim_{x\to-\infty}\frac{\sqrt{x^{4}\left(4+\frac{3}{x^{2}}+\frac{1}{x}\right)}}{x^{2}\left(\frac{1}{x^{2}}-2\right)}\\ & =\lim_{x\to-\infty}\frac{\sqrt{x^{4}}\sqrt{4+\frac{3}{x^{2}}+\frac{1}{x}}}{x^{2}\left(\frac{1}{x^{2}}-2\right)}\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}\left(\frac{1}{x^{2}}-2\right)} & =\lim_{x\to-\infty}\frac{\bcancel{x^{2}}\sqrt{4+\frac{3}{x^{2}}+\frac{1}{x}}}{\bcancel{x^{2}}\left(\frac{1}{x^{2}}-2\right)}\\ & =\lim_{x\to-\infty}\frac{\sqrt{4+\frac{3}{x^{2}}+\frac{1}{x}}}{\left(\frac{1}{x^{2}}-2\right)}\\ & =\frac{\lim_{x\to-\infty}\sqrt{4+\frac{3}{x^{2}}+\frac{1}{x}}}{\lim_{x\to-\infty}\left(\frac{1}{x^{2}}-2\right)}\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}\left(\frac{1}{x^{2}}-2\right)} & =\frac{\sqrt{\lim_{x\to-\infty}\left(4+\frac{3}{x^{2}}+\frac{1}{x}\right)}}{\lim_{x\to-\infty}\left(\frac{1}{x^{2}}-2\right)}\\ & =\frac{\sqrt{4+3(0)+0}}{0-2}\\ & =\frac{2}{-2}\\ & =-1.\end{aligned}

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

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

Example 10

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

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

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}}{3t-9} & =\lim_{t\to\infty}\frac{\sqrt{t^{2}\left(1+\dfrac{6}{t^{2}}\right)}}{t\left(3-\dfrac{9}{t}\right)}\\ & =\lim_{t\to\infty}\frac{\sqrt{t^{2}}\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)}\end{aligned} Recall that $$\sqrt{t^{2}}=|t|$$. So $\lim_{t\to\infty}\frac{\sqrt{t^{2}}\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)}=\lim_{t\to\infty}\frac{|t|\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)}$ Because when $$t\to\infty$$, the values under consideration are positive $$t>0$$, \begin{aligned} \lim_{t\to\infty}\frac{|t|\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)} & =\lim_{t\to\infty}\frac{\bcancel{t}\sqrt{1+\dfrac{6}{t^{2}}}}{\bcancel{t}\left(3-\dfrac{9}{t}\right)}\\ & =\frac{\lim_{t\to\infty}\sqrt{1+\dfrac{6}{t^{2}}}}{\lim_{t\to\infty}\left(3-\dfrac{9}{t}\right)}\\ & =\frac{\sqrt{\lim_{t\to\infty}\left(3+\dfrac{6}{t^{2}}\right)}}{\lim_{t\to\infty}3-\lim_{t\to\infty}\dfrac{9}{t}}\\ & =\frac{\sqrt{\lim_{t\to\infty}1+\lim_{t\to\infty}\dfrac{6}{t}}}{\lim_{t\to\infty}3-\lim_{t\to\infty}\dfrac{9}{t}}\\ & =\frac{\sqrt{1+0}}{3-0}\\ & =\frac{1}{3}.\end{aligned} Here we have used the fact that $$\frac{1}{t^{r}}\to0$$ 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}}{3t-9} & =\lim_{t\to-\infty}\frac{\sqrt{t^{2}}\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)}\\ & =\lim_{t\to-\infty}\frac{|t|\sqrt{1+\dfrac{6}{t^{2}}}}{t\left(3-\dfrac{9}{t}\right)}\\ & =\lim_{t\to-\infty}\frac{-\bcancel{t}\sqrt{1+\dfrac{6}{t^{2}}}}{\bcancel{t}\left(3-\dfrac{9}{t}\right)}\\ & =\frac{-\sqrt{\lim_{t\to\infty}1+\lim_{t\to\infty}\dfrac{6}{t^{2}}}}{\lim_{t\to-\infty}3-\lim_{t\to\infty}\dfrac{9}{t}}\\ & =\frac{-\sqrt{1+0}}{3-0}\\ & =-\frac{1}{3}.\end{aligned}

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

 Figure: Graph of $$f(t)=\frac{\sqrt{t^{2}+6}}{3t-9}$$. 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 11

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

Solution

Similar to the previous examples, we factor out the highest power of $$x$$ inside the radical: $\lim_{x\to-\infty}\left(\sqrt{x^{4}+5}-2x^{2}\right)=\lim_{x\to-\infty}\left(\sqrt{x^{4}\left(1+\frac{5}{x^{4}}\right)}-2x^{2}\right)$ Because $$x^{4}\geq0$$ and $$1+\frac{5}{x^{4}}\geq0$$, thus $\sqrt{x^{4}\left(1+\frac{5}{x^{4}}\right)}=\sqrt{x^{4}}\sqrt{1+\dfrac{5}{x^{4}}}=x^{2}\sqrt{1+\dfrac{5}{x^{4}}}$ and \begin{aligned} \lim_{x\to-\infty}\left(\sqrt{x^{4}\left(1+\frac{5}{x^{4}}\right)}-2x^{2}\right) & =\lim_{x\to-\infty}\left(x^{2}\sqrt{1+\frac{5}{x^{4}}}-2x^{2}\right)\\ & =\lim_{x\to-\infty}\left(x^{2}\left(\sqrt{1+\frac{5}{x^{4}}}-2\right)\right)\\ & =\left(\lim_{x\to-\infty}x^{2}\right)\lim_{x\to-\infty}\left(\sqrt{1+\frac{5}{x^{4}}}-2\right)\\ & =\left(\lim_{x\to-\infty}x^{2}\right)\left(\sqrt{\lim_{x\to-\infty}1+\lim_{x\to-\infty}\frac{5}{x^{4}}}-\lim_{x\to-\infty}2\right)\\ & =(+\infty)(\sqrt{1+0}-2)\\ & =-\infty\end{aligned} The graph of $$f(x)=\sqrt{x^{4}+5}-2x^{2}$$ is shown below.

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

Example 12

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

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: $\lim_{x\to-\infty}\left(\sqrt{x^{4}+5}-x^{2}\right)=\lim_{x\to-\infty}\left(x^{2}\left(\sqrt{1+\frac{5}{x^{4}}}-1\right)\right)$ However, because $\lim_{x\to-\infty}x^{2}=\infty,\quad\lim_{x\to-\infty}\left(\sqrt{1+\frac{5}{x^{4}}}-1\right)=0$ 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 $$\sqrt{x^{4}+5}+x^{2}$$: $\lim_{x\to-\infty}\left(\sqrt{x^{4}+5}-x^{2}\right)=\lim_{x\to-\infty}\left(\left(\sqrt{x^{4}+5}-x^{2}\right)\frac{\sqrt{x^{4}+5}+x^{2}}{\sqrt{x^{4}+5}+x^{2}}\right)$ Recall that $$(A-B)(A+B)=A^{2}-B^{2}$$. Therefore,

\begin{aligned}
\small{\lim_{x\to-\infty}\left(\frac{\left(\sqrt{x^{4}+5}-x^{2}\right)\left(\sqrt{x^{4}+5}+x^{2}\right)}{\sqrt{x^{4}+5}+x^{2}}\right) }& =\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}}\6pt] & =\lim_{x\to-\infty}\frac{5}{x^{2}\left(\sqrt{1+\frac{5}{x^{4}}}+1\right)}\\[6pt] & =\small{\frac{\lim_{x\to-\infty}5}{\displaystyle{\lim_{x\to-\infty}}x^{2}\left(\sqrt{\displaystyle{\lim_{x\to-\infty}}1+\displaystyle{\lim_{x\to-\infty}}\frac{5}{x^{4}}}+\displaystyle{\lim_{x\to-\infty}}1\right)}}\\[6pt] & =\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 $$f(x)\to0$$ as $$x\to\pm\infty$$. Example 13 Find \[ \lim_{x\to-\infty}\left(\sqrt{x^{4}+5x^{2}}-x^{2}\right)

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}\left(\sqrt{x^{4}+5x^{2}}-x^{2}\right) & =\lim_{x\to-\infty}\left(x^{2}\sqrt{1+\frac{5}{x^{2}}}-x^{2}\right)\\ & =\lim_{x\to-\infty}\left(x^{2}\left(\sqrt{1+\frac{5}{x^{2}}}-1\right)\right)\end{aligned} and $$\lim_{x\to-\infty}x^{2}=\infty$$ and $$\lim_{x\to-\infty}\left(\sqrt{1+\frac{5}{x^{2}}}-1\right)=0$$. So like the previous example, let’s multiply and divide by the conjugate of the given function. \begin{aligned} \lim_{x\to-\infty}\left(\sqrt{x^{4}+5x^{2}}-x^{2}\right) & =\lim_{x\to-\infty}\left(\left(\sqrt{x^{4}+5x^{2}}-x^{2}\right)\frac{\sqrt{x^{4}+5x^{2}}+x^{2}}{\sqrt{x^{4}+5x^{2}}+x^{2}}\right)\\ & =\lim_{x\to-\infty}\frac{x^{4}+5x^{2}-x^{4}}{\sqrt{x^{4}+5x^{2}}+x^{2}}\\ & =\lim_{x\to-\infty}\frac{5\bcancel{x^{2}}}{\bcancel{x^{2}}\left(\sqrt{1+\frac{5}{x^{2}}}+1\right)}\\ & =\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 $$f(x)=\sqrt{x^{4}+5x^{2}}-x^{2}$$ is shown below.

 Figure: Graph of $$f(x)=\sqrt{x^{4}+5x^{2}}-x^{2}$$. It is clear from this graph that $$f(x)\to2.5$$ as $$x\to\pm\infty$$.

Example 14

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

Solution

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