The limit laws

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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 $x^b$ 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 A rational function h, meaning a function of the form $h(x)=\frac {f(x)}{g(x)}$ where $f$ and $g$ are polynomials, 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 $b = \lim _{x\to a} g(x)$ , 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 $x=a$ , and if $f$ is continuous at $g(a)$ , then $f(g(x))$ is continuous at $x=a$ .

Compute the following limit using limit laws: $\lim _{x \to 0} \sqrt {\cos (x)}$

We will use the Composition Limit Law. Let $f(g(x))=\sqrt {\cos {x}}=(\cos {x})^{\frac {1}{2}}$ where $f(x)=\sqrt {x}=x^{\frac {1}{2}}$ , and $g(x)=\cos {x}$ . Now, continuity of cosine implies that $\lim _{x \to 0}g(x)=\lim _{x \to 0} \cos {x}=\cos {0}=1$ . Continuity of $x^b$ , where $b=1/2$ , implies that $f$ is continuous at $1=\lim _{x \to 0}g(x)$ . Now, the Composition Limit Law implies that $\lim _{x \to 0} \sqrt {\cos {x}}= \sqrt {\lim _{x \to 0} \cos {x}}=\sqrt {1}=1$

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

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.

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? $\lim _{x\to 0} x\sin (1/x)$

yes no

If we are trying to use limit laws to compute this limit, we would first have to use the Product Law to say that $\lim _{x\to 0}x\sin (1/x)= \lim _{x\to 0} x \cdot \lim _{x\to 0} \sin (1/x).$ We are only allowed to use this law if both limits exist, so we must check this first. We know from continuity that $\lim _{x\to 0}x=0.$ However, we also know that $\sin (1/x)$ oscillates “wildly” as $x$ approaches $0$ , and so the limit $\lim _{x\to 0} \sin (1/x)$ does not exist. Therefore, we cannot use the Product Law.

Can this limit be directly computed by limit laws? $\lim _{x\to 0} \cot (x^3)$

yes no

Notice that $\cot (x^3) = \frac {\cos (x^3)}{\sin (x^3)}.$ If we are trying to use limit laws to compute this limit, we would like to use the Quotient Law to say that $\lim _{x\to 0} \frac {\cos (x^3)}{\sin (x^3)} = \frac {\lim _{x\to 0} \cos (x^3)}{\lim _{x\to 0} \sin (x^3)}.$ We are only allowed to use this law if both limits exist and the denominator is not $0$ . We suspect that the limit in the denominator might equal $0$ , so we check this limit. $\begin{align*} \lim_{x\to 0} \sin(x^3) &= \sin(\lim_{x\to 0}x^3)\\ &=\sin(0) \\ &=0. \end{align*}$ This means that the denominator is zero and hence we cannot use the Quotient Law.

Can this limit be directly computed by limit laws? $\lim _{x\to 0}\sec ^2(e^x-1)$

yes no

Compute: $\lim _{x\to 0}\sec ^2(e^x-1)\begin {prompt} =\answer {1}\end {prompt}$

Notice that $\lim _{x\to 0} \sec ^2(e^x-1) = \lim _{x\to 0} \frac {1}{\cos ^2(e^x-1)}.$ If we are trying to use Limit Laws to compute this limit, we would now have to use the Quotient Law to say that $\lim _{x\to 0} \frac {1}{\cos ^2(e^x-1)} = \frac { \lim _{x\to 0}1}{ \lim _{x\to 0}\cos ^2(e^x-1)}.$ We are only allowed to use this law if both limits exist and the denominator is not $0$ . Let’s check the denominator and numerator separately. First we’ll compute the limit of the denominator: $\begin{align*} \lim_{x\to 0}\cos^2(e^x-1) &= (\lim_{x\to 0}\cos{(e^x-1)})^{2}\\ &= \Bigl(\cos{\Big(\lim_{x\to 0}(e^x-1)\Bigr)}\Bigr)^{2}\\ & =\cos^2\Bigl(\lim_{x\to 0}(e^x-1)\Bigr)\\ &=\cos^2\Bigl(\lim_{x\to 0}(e^x)-\lim_{x\to 0}(1)\Bigr)\\ &=\cos^2(1-1)\\ &= \cos^2(0)\\ &=1 \end{align*}$ Therefore, the limit in the denominator exists and does not equal $0$ . We can use the Quotient Law, so we will compute the limit of the numerator: $\lim _{x\to 0}1=1$ Hence $\frac { \lim _{x\to 0}1}{ \lim _{x\to 0}\cos ^2(e^x-1)} = \frac {1}{1}=1$

Can this limit be directly computed by limit laws? $\lim _{x\to 1}{(x-1)\cdot \csc (\ln (x))}$

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 $\lim _{x\to 1}{(x-1)\cdot \csc (\ln (x))}= \lim _{x\to 1}{(x-1)\cdot \lim _{x\to 1}\csc (\ln (x))}.$ We are only allowed to use this law if both limits exist. Let’s check each limit separately. $\begin{align*} \lim_{x\to 1} (x-1) &= \lim_{x\to 1} (x)-\lim_{x\to 1}(1)\\ &=1-1\\ &=0. \end{align*}$ So this limit exists. Now we check the other factor. Notice that $\lim _{x\to 1}\csc (\ln (x)) = \frac {1}{\sin (\ln (x))}.$ If we are trying to use limit laws to compute this limit, we would now have to use the Quotient Law to say that $\frac {1}{\sin (\ln (x))} = \frac {\lim _{x\to 1}1}{\lim _{x\to 1}\sin (\ln (x))}.$ We are only allowed to use this law if both limits exist and the denominator does not equal $0$ . The limit in the numerator definitely exists, so let’s check the limit in the denominator. $\begin{align*} \lim_{x\to 1}\sin(\ln(x)) &= \sin(\lim_{x\to 1}\ln(x))\\ &=\sin(\ln(1))\\ &= \sin(0)\\ &= 0 \end{align*}$ Since the denominator is $0$ , we cannot apply the Quotient Law.

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.