The limit laws

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\CancelColor }[0]{} \newcommand {\cancelto }[0]{1} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$ $% This file was *autogenerated* from digInLimitLaws.sagetex.sage with % sagetex.py version 2015/08/26 v3.0-92d9f7a %4dd36c5d1eeb3d7f31c084fde3511a48% md5sum of corresponding .sage file (minus "goboom","current_tex_line",and pause/unpause lines)$

We give basic laws for working with limits.

In this section, we present a handful of rules called the Limit Laws that allow us to find limits of various combinations of functions.

Limit Laws Suppose that $\lim _{x\to a}f(x)=L$ , $\lim _{x\to a}g(x)=M$ .

Sum/Difference Law: $\lim _{x\to a} (f(x) \pm g(x)) = \lim _{x\to a}f(x) \pm \lim _{x\to a}g(x)=L \pm M$ .
Product Law: $\lim _{x\to a} (f(x)g(x)) = \lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x)=LM$ .
Quotient Law: $\lim _{x\to a} \frac {f(x)}{g(x)} = \frac {\lim _{x\to a}f(x)}{\lim _{x\to a}g(x)}=\frac {L}{M}$ , if $M\ne 0$ .

True or false: If $f$ and $g$ are continuous functions on an interval $I$ , then $f\pm g$ is continuous on $I$ .

True False

This follows from the Sum/Difference Law.

True or false: If $f$ and $g$ are continuous functions on an interval $I$ , then $f/g$ is continuous on $I$ .

True False

In this case, $f/g$ will not be continuous for $x$ where $g(x) = 0$ .

Compute the following limit using limit laws: $\lim _{x\to 1}(5x^2+3x-2)$

Well, get out your pencil and write with me: $\lim _{x\to 1} (5x^2+3x-2) = \lim _{x\to 1} 5x^2 + \lim _{x\to 1} \answer [given]{3x} - \lim _{x\to 1}2$ by the Sum/Difference Law. So now $= 5\lim _{x\to 1} x^2 + 3\lim _{x\to 1} x - \lim _{x\to 1}\answer [given]{2}$ by the Product Law. Finally by continuity of $x^k$ and $k$ , $= 5(1)^2 + 3(1) - 2 =\answer [given]{6}.$ We can check our answer by looking at the graph of $y=f(x)$ : $\graph {5x^2+3x-2}$

We can generalize the example above to get the following theorems.

Continuity of Polynomial Functions All polynomial functions, meaning functions of the form $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ where $n$ is a whole number and each $a_i$ is a real number, are continuous for all real numbers.

Continuity of Rational Functions Let $f$ and $g$ be polynomials. Then a rational function, meaning an expression of the form $h=\frac {f}{g}$ is continuous for all real numbers except where $g(x)=0$ . That is, rational functions are continuous wherever they are defined.

Let $a$ be a real number such that $g(a)\neq 0$ . Then, since $g(x)$ is continuous at $a$ , $\lim _{x\to a} g(x) \neq 0$ . Therefore, write with me, $\lim _{x \to a} h(x) = \lim _{x\to a} \frac {f(x)}{g(x)}$ and now by the Quotient Law, $\frac {\lim _{x\to a} f(x)}{ \lim _{x\to a} g(x)}$ and by the continuity of polynomials we may now set $x=a$ $\frac {f(a)}{g(a)}=h(a).$ Since we have shown that $\lim _{x\to a} h(x) = h(a)$ , we have shown that $h$ is continuous at $x=a$ .

Where is $f(x) = \frac {x^2-3x+2}{x-2}$ continuous?

for all real numbers at $x=2$ for all real numbers, except $x=2$ impossible to say

Now, we give basic rules for how limits interact with composition of functions.

Composition Limit Law If $f(x)$ is continuous at $x = \lim _{x\to a} g(x)$ , then $\lim _{x\to a} f(g(x)) = f(\lim _{x\to a} g(x)).$

Because the limit of a continuous function is the same as the function value, we can now pass limits inside continuous functions.

Continuity of Composite Functions If $g$ is continuous at $x=a$ , then $f(g(x))$ is continuous at $x=a$ .

Compute the following limit using limit laws: $\lim _{x \to e} \sqrt {\ln (x)}$

By continuity of $x^k$ , assuming $\lim _{x \to e} \ln (x) >0$ , $\lim _{x \to e} \sqrt {\ln (x)} = \sqrt {\lim _{x \to e} \ln (x)},$ and now since the natural logarithm function is continuous for all $x>0$ , $\begin{align*} \sqrt{\lim_{x \to e} \ln(x)} & =\sqrt{\ln(e)}\\ &= \sqrt{1}\\ &= 1. \end{align*}$

We can confirm our results by checking out the graph of $y=f(x)$ : $\graph {\sqrt {\ln (x)}}$

Many of the Limit Laws and theorems about continuity in this section might seem like they should be obvious. You may be wondering why we spent an entire section on these theorems. The answer is that these theorems will tell you exactly when it is easy to find the value of a limit, and exactly what to do in those cases.

The most important thing to learn from this section is whether the limit laws can be applied for a certain problem, and when we need to do something more interesting. We will begin discussing those more interesting cases in the next section. For now, we end this section with a question:

A list of questions

Let’s try this out.

Can this limit be directly computed by limit laws? $\lim _{x\to 2}\frac {x^2+3x+2}{x+2}$

yes no

Compute: $\lim _{x\to 2}\frac {x^2+3x+2}{x+2}\begin {prompt} =\answer {3}\end {prompt}$

Since $f(x)=\frac {x^2+3x+2}{x+2}$ is a rational function, and the denominator does not equal $0$ , we see that $f(x)$ is continuous at $x=2$ . Thus, to find this limit, it suffices to plug $2$ into $f(x)$ .

Can this limit be directly computed by limit laws? $\lim _{x\to 2}\frac {x^2-3x+2}{x-2}$

yes no

$f(x) = \frac {x^2-3x+2}{x-2}$ is a rational function, but the denominator $x-2$ equals $0$ when $x=2$ . None of our current theorems address the situation when the denominator of a fraction approaches $0$ .

Can this limit be directly computed by limit laws? $\lim _{x\to 0} x\ln x$

yes no

If we are trying to use limit laws to compute this limit, we would have to use the Product Law to say that $\lim _{x\to 0} x\ln x =\lim _{x\to 0} x \cdot \lim _{x\to 0}\ln x.$ We are only allowed to use this law if both limits exist. We know $\lim _{x\to 0} x = 0$ , but what about $\lim _{x\to 0}\ln x$ ? We do not know how to find $\lim _{x\to 0}\ln x$ using limit laws because $0$ is not in the domain of $\ln x$ .

Can this limit be directly computed by limit laws? $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}}$

yes no

Compute: $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}}\begin {prompt} =\answer {0}\end {prompt}$

If we are trying to use limit laws to compute this limit, we would have to use the Quotient Law to say that $\lim _{x\to 0}\frac {2^x-1}{3^{x-1}} = \frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})}.$ We are only allowed to use this law if both limits exist and the denominator does not equal $0$ . Let’s check each limit separately, starting with the denominator $\begin{align*} \lim_{x\to 0}(3^{x-1}) &=3^{\lim_{x\to0}(x-1)}\\ &=3^{-1}\\ &=\frac{1}{3} \end{align*}$

On the other hand the limit in the numerator is

$\begin{align*} \lim_{x\to 0}(2^x-1) &=\lim_{x\to0}(2^x)-\lim_{x\to0}(1)\\ &=1-1\\ &=0 \end{align*}$

The limits in both the numerator and denominator exist and the limit in the denominator does not equal $0$ , so we can use the Quotient Law. We find: $\frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})} =\frac {0}{\frac {1}{3}}=0.$

Can this limit be directly computed by limit laws? $\lim _{x\to 0}(1+x)^{1/x}$

yes no

We do not have any limit laws for functions of the form $f(x)^{g(x)}$ , so we cannot compute this limit.