2.3: Limits Algebraically

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below are two videos showing worked examples.

Example 1:

Example 2:

Problems

We have the following limit laws:

Limit Laws Assume $\lim _{x \to a} f(x)$ and $\lim _{x \to a} g(x)$ exist. Then:

(Add./Sub.) $\lim _{x \to a} [f(x)\pm g(x)] = \lim _{x \to a} f(x) \pm \lim _{x \to a} g(x)$ .
(Constant) $\lim _{x \to a} c\cdot g(x) =c \lim _{x \to a} g(x)$ .
(Mult.) $\lim _{x \to a} f(x)g(x) = \lim _{x \to a} f(x) \cdot \lim _{x \to a} g(x)$ .
(Div.) $\displaystyle \lim _{x \to a} \frac {f(x)}{g(x)} = \frac {\lim _{x \to a} f(x)}{\lim _{x \to a} g(x)}$ (if $\lim _{x \to a} g(x) \neq 0$ )
(Power) $\lim _{x \to a} [f(x)]^n = \left (\lim _{x \to a} f(x) \right )^n$
(Root) $\lim _{x \to a} \sqrt [n]{f(x)} = \sqrt [n]{\lim _{x \to a} f(x)}$

Use the limit laws to evaluate $\lim _{x \to 3} \frac {x^2}{x+1}$ .

We can use limit laws to calculate this. The steps are indicated below. Pick the law that is being used in each step.

$\begin{eqnarray} \lim_{x \to 3} x^2 \sqrt{x+1} &=&\left[ \lim_{x \to 3} x^2 \right]\left[ \lim_{x \to 3} \sqrt{x+1} \right] \\ &=& \left[\lim_{x \to 3} x \right]^2 \left[\lim_{x \to 3} \sqrt{x+1} \right] \\ &=& \left[\lim_{x \to 3} x \right]^2 \sqrt{\lim_{x \to 3} (x+1)} \\ &=&\left[\lim_{x \to 3} x \right]^2 \sqrt{\lim_{x \to 3} x+ \lim_{x \to 3} 1} \\ &=& 3^2 \cdot (3+1) \\ &=& \answer{36}.\\ \end{eqnarray}$

(a): Step 1: Limit law Add./Sub.ConstantMult.Div.PowerRoot
(b): Step 2: Limit law Add./Sub.ConstantMult.Div.PowerRoot
(c): Step 3: Limit law Add./Sub.ConstantMult.Div.PowerRoot
(d): Step 4: Limit law Add./Sub.ConstantMult.Div.PowerRoot

Here, we will see how the product limit law doesn’t necessarily apply if at least one of the individual limits do not exist. Consider $f(x) = x$ and $g(x) = \frac {1}{x}$ , and let $a=0$ . Then we can say $\lim _{x \to 0} f(x) = \answer {0},$ and $\lim _{x \to 0} g(x) = \answer {DNE}.$ If we were going to use the product limit law freely, we might be tempted to write that $\lim _{x \to 0} f(x)g(x) = 0,$ since the limit of $f(x)$ is $0$ . We will now see this is not the case. Notice that, as long as $x \neq 0$ , we can say the product $f(x)g(x)$ is equal to $f(x)g(x) = \answer {1} \quad \text { for } x \neq 0.$ But this means that $\lim _{x \to 0} f(x)g(x) = \lim _{x \to 0} 1 = \answer {1}.$ So we see we get a limit which is not zero, showing that the limit law could not have been applied since the limit of $g(x)$ did not exist.

If $f(x)$ is a polynomial or a rational function (i.e. a quotient of polynomials), and $x=a$ is in the domain of $f(x)$ , then to evaluate $\lim _{x \to a} f(x)$ , we can just plug $x=a$ into the function (i.e. the limit equals $f(a)$ ).

Use the fact above to evaluate $\lim _{x \to 1} \frac {x^2+1}{x+1}$ .

Notice that $\displaystyle f(x) =\frac {x^2+1}{x+1}$ is a rational function. The point $x=1$ isis not in the domain of $f(x)$ , which means we can evaluate the limit by plugging $x=1$ into $f(x)$ . Therefore, $\lim _{x \to 1} \frac {x^2+1}{x+1} = \answer {1}.$

Notice $f(1) = 1$ .

If $f(x) = \begin {cases} x^2+1, & \text {if } x >1 \\ 3-2x, & \text {if } x \leq 1 \end {cases}$ evaluate $\lim _{x \to 1} f(x)$ if it exists.

Notice that this function has two different definitions for the left and right of $x=1$ , so as we approach $x=1$ from the left and from the right, we will need to use a different piece of the function. On the left of $x=1$ , the function is equal to $x^2+1$ $3-2x$ , and so $\lim _{x \to 1^-} f(x) = \lim _{x \to 1^-} \answer {3-2x}.$ Now, $3-2x$ is a polynomial, so we can just plug in $x=1$ into this polynomial to find $\lim _{x \to 1^-} f(x) = \lim _{x \to 1^-} (3-2x)= \answer {1}.$ On the right of $x=1$ , the function is equal to $x^2+1$ $3-2x$ , and so $\lim _{x \to 1^+} f(x) = \lim _{x \to 1^+} \answer {x^2+1}.$ Again, $x^2+1$ is a polynomial, so we can just plug in $x=1$ into this polynomial to find $\lim _{x \to 1^+} f(x) = \lim _{x \to 1^+} (x^2+1)= \answer {2}.$ Since the left limit does not equal the right limit, we can say the limit does not exist.

The question to address now is the following:

What happens if $x=a$ is not in the domain of the function of the rational function?

Calculate $\lim _{x \to 2} \frac {x^2-4}{x^2-3x+2}$ .

The function $f(x) = \frac {x^2-4}{x^2-3x+2}$ is a rational function, but $x=2$ is not in the domain of this function since the denominator becomes $0$ when you plug in $x=2$ . However, in this case, so does the numerator. If you get $0/0$ when you plug in the point, what that tells you is you have to perform some algebra on the function to help you determine what is happening. Here, for example, you can factor the numerator and denominator. Notice $f(x) = \frac {(x-2)(\answer {x+2})}{(x-2)(\answer {x-1})},$ and the $x-2$ terms will cancel, leaving $f(x) = \frac {x+2}{x-1}.$ Notice that you can plug in $x=2$ into this rational expression, which means $\lim _{x \to 2} \frac {x^2-4}{x^2-3x+2} = \lim _{x \to 2} \frac {x+2}{x-1} = \answer {4}.$

What cancelling the $x-2$ ’s tells us is that the function $\frac {x^2-4}{x^2-3x+2}$ is the same as the function $\frac {x+2}{x-1}$ , just with a hole at $x=2$ (since $x=2$ is not in the domain of the original function).
If, when you plug in your point, you get $0/0$ , you must perform more algebra to see if the limit exists.

The limit $\lim _{x \to 0} \frac {x}{x^2-x} = \answer {-1}$ .

When you plug in $x=0$ into $f(x) = x/(x^2-x)$ , you get $0/0$ , which means you have to do more algebra. Here, we can factor the denominator, giving us $f(x) = \frac {x}{x(\answer {x-1})} = \frac {1}{\answer {x-1}}.$ Therefore, $\lim _{x \to 0} \frac {x}{x^2-x} = \lim _{x \to 0} \frac {1}{x-1} = \answer {-1}.$

Evaluate $\lim _{x \to 9} \frac {x-9}{\sqrt {x}-3}$ .

Again, we get $0/0$ when we plug in $x=9$ , so we must perform some algebra. We could factor here, but we will demonstrate another method. One useful trick is known as “multiplying by the conjugate.” In this case, we want to remove the square root on the bottom, so we will multiply the top and bottom of the fraction by $\sqrt {x}+3$ (the conjugate of $\sqrt {a}+\sqrt {b}$ is $\sqrt {a}-\sqrt {b}$ , and vice versa). To see why this is useful, notice that the product expands to $(\sqrt {x}-3)(\sqrt {x}+3) = \answer {x-9},$ removing the square root for us. Performing this operation gives $\frac {x-9}{\sqrt {x}-3} = \frac {x-9}{\sqrt {x}-3} \cdot \frac {\sqrt {x}+3}{\sqrt {x}+3} = \frac {(x-9)(\sqrt {x}+3)}{x-9}.$ Notice we do not expand the product in the numerator since this would not simplify the expression for us. Now, the $x-9$ ’s can cancel, and we are left only with $\frac {(x-9)(\sqrt {x}+3)}{x-9} =\answer { \sqrt {x}+3}.$ Therefore, $\lim _{x \to 9} \frac {x-9}{\sqrt {x}-3} = \lim _{x \to 9} \sqrt {x}+3 = \answer {6}.$

The limit $\lim _{x \to 1} \frac {\sqrt {x^2+3}-2}{x-1} = \answer {1/2}$ .

Multiply the numerator and denominator by $\sqrt {x^2+3}+2$ , and then you’ll need to factor.

The answer is $1/2$ .

Compute $\lim _{x \to 3^+} \frac {2x}{x-3}$ .

Notice, here, we do not get $0/0$ , we get $6/0$ , which tells us that the limit will not exist, and instead we will have an asymptote. If we wanted to see whether the limit was $+ \infty$ or $- \infty$ , we could do some quick investigative work. As $x$ approaches $3$ from the right, the numerator will just approach $6$ , and the denominator will be a very small positivenegative number. When you divide $6$ by a really small number (think .00000001), the result will be a really bigreally small positivenegative number, which means the limit is going to be $+\infty$ .

If $\displaystyle f(x) = \frac {x}{(x+1)^2}$ , then we can say

$\lim _{x \to -1^-} f(x) = \answer {-\infty }$
$\lim _{x \to -1^+} f(x) = \answer {-\infty }$
$\lim _{x \to -1} f(x) = \answer {-\infty }$
$\lim _{x \to 0} f(x) =\answer {0}$ .

When you plug in the first three values, you get $-1/0$ , meaning we have an asymptote. However, for the last one, notice that $x=0$ is in the domain of $f(x)$ .

We have one more theorem to help us compute limits.

Squeeze Theorem Suppose $h(x) \leq f(x) \leq g(x)$ near $x=a$ , and suppose $\lim _{x \to a} h(x) = \lim _{x \to a} g(x) = L$ . Then $\lim _{x \to a} f(x) = L$ .

When using the squeeze theorem to compute limits, your given function should be the term sandwiched between the other two.

Consider $\lim _{x \to \pi } \sin ^2(x)\sin \left (\frac {1}{x-\pi } \right )$ .

Notice that if you were to plug in $x = \pi$ , the $\sin ^2(x)$ term would be $0$ , but the other term is undefined because of the $\displaystyle \frac {1}{x-\pi }$ . Therefore, we cannot use the limit laws directly, and this is a classic squeeze theorem problem. In this case, our problem term is $\sin (1/(x-\pi ))$ . Since $\sin$ of anything is between $-1$ and $1$ , we can definitely say $-1 \leq \sin \left (\frac {1}{x-\pi } \right ) \leq 1.$ Now that we have sandwiched our problem term, we will try to make the middle look like our original function. To do this, let’s multiply by $\sin ^2(x)$ everywhere, and we get $-\sin ^2(x) \leq \sin ^2(x)\sin \left (\frac {1}{x-\pi } \right ) \leq \answer {\sin ^2(x)}.$ Notice, $\lim _{x \to \pi } -\sin ^2(x) = \answer {0},$ $\lim _{x \to \pi } \sin ^2(x) = \answer {0}.$ Therefore, by the Squeeze Theorem, we can say $\lim _{x \to \pi } \sin ^2(x)\sin \left (\frac {1}{x-\pi } \right ) = \answer {0}.$

The limit $\lim _{x \to 0^+} x\cos (x)\cos (1/x) = \answer {0}$ .

This is another Squeeze Theorem problem because of the $\cos (1/0)$ term. In this case, the $\cos (1/x)$ term is the problem term, so we bound that one. $-1 \leq \cos (1/x) \leq 1.$ Now, we transform these inequalities so that the middle term is our original function. In this case, we can achieve this by multiplying everything by $\answer {x\cos (x)}$ . Doing this gives $\answer {-x\cos (x)} \leq x \cos (x) \cos (1/x) \leq \answer {x\cos (x)}.$ Notice $\lim _{x \to 0^+} -x\cos (x) = \lim _{x \to 0^+} x\cos (x) = \answer {0}.$ Therefore, by Squeeze Theorem, our original limit is $\answer {0}$ .

Suppose $f(x)$ is a function such that, for $-4 \leq x \leq 4$ , we have $(x-3)^2+1 \leq f(x) \leq |x-3|+1.$ Then we can say that $\lim _{x \to 3} f(x) = \answer {1}.$

What does Squeeze Theorem say?