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 \[ \begin{aligned} \lim_{x\to\infty}\left(\frac{2x^{3}}{x^{2}-4}-\frac{2x^{2}}{x-1}\right) & =\lim_{x\to\infty}\frac{-2x^{3}+8x^{2}}{x^{3}-x^{2}-4x+4}\\ & =\lim_{x\to\infty}\frac{-2x^{3}}{x^{3}}\\ & =-2.\end{aligned} \]

Limits Involving Radicals

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.