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 $\displaystyle \lim _{x\to a}f(x)=L$ , $\displaystyle \lim _{x\to a}g(x)=M$ , where $L$ and $M$ are real numbers.

Constant Multiple Law: $\displaystyle \lim _{x\to a} kf(x) = k\displaystyle \lim _{x\to a}f(x)=kL$ .
Sum/Difference Law: $\displaystyle \lim _{x\to a} (f(x) \pm g(x)) = \lim _{x\to a}f(x) \pm \lim _{x\to a}g(x)=L \pm M$ .
Product Law: $\displaystyle \lim _{x\to a} (f(x)g(x)) = \displaystyle \lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x)=LM$ .
Quotient Law: $\displaystyle \lim _{x\to a} \frac {f(x)}{g(x)} = \frac {\displaystyle \lim _{x\to a}f(x)}{\displaystyle \lim _{x\to a}g(x)}=\frac {L}{M}$ , if $M\ne 0$ .
Power Law: If $n$ is a positive whole number, $\displaystyle \lim _{x \to a} f(x)^n = L^n.$
Root Law: If $n$ is a positive whole number, $\displaystyle \lim _{x \to a} \sqrt [n]{f(x)} = \sqrt [n]{L}$
( $L$ is required to be positive if $n$ is even.)

Notice that this theorem starts by supposing that $\displaystyle \lim _{x\to a}f(x)$ and $\displaystyle \lim _{x\to a}g(x)$ exist. Without this supposition, none of the limit laws are necessarily true! Be careful not to use a limit law unless both $\displaystyle \lim _{x\to a}f(x)$ and $\displaystyle \lim _{x\to a}g(x)$ exist.

Hopefully you noticed that the Quotient Law has the additional requirement that the limit of the denominator, $\displaystyle \lim _{x\to a}g(x)= M \neq 0$ . You can’t divide by $0$ , after all.

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

Well, get out your pencil and write with me: $\displaystyle \lim _{x\to 1} (5x^2+3x-2) = \lim _{x\to 1} 5x^2 + \lim _{x\to 1} \answer [given]{3x} - \lim _{x\to 1}2$ by the Sum/Difference Law. So now by the Constant Multiple Law followed by the Power Law, $= 5\displaystyle \lim _{x\to 1} x^2 + 3\lim _{x\to 1} x - \lim _{x\to 1}\answer [given]{2}$ $= 5\left (\displaystyle \lim _{x\to 1} x\right )^2 + 3\lim _{x\to 1} x - \lim _{x\to 1}\answer [given]{2}$ $= 5(1)^2 + 3(1) - 2 =\answer [given]{6}.$ Ultimately, the Limit Laws allowed us to make short work of evaluating this limit. Notice that once we used all the Limit Laws we possibly could, we were left to evaluate $\displaystyle \lim _{x\to 1} x$ . Luckily, this is an example of a very ‘nice’ kind of function that we can evaluate simply by plugging in $1$ for $x$ . In other words, if we say $h(x) = x$ , then $\displaystyle \lim _{x\to 1} h(x) = h(1) = 1$ so the value of the limit at $x=1$ is the same as the function value at $x=1$ . This is precisely the condition that makes this function so ’nice’!

We can check our answer by looking at the graph of $y=f(x)$ . Pause for a second to make sure you are able to confirm that our answer seems correct using the graph below. $\graph {5x^2+3x-2}$

The most important thing to learn from this section is whether the limit laws can be applied for a certain problem, or if we need to do something more interesting. We will begin discussing those more interesting cases in the next section. For now, let’s practice applying the limits laws to evaluate limits.

When can you use the limit laws?

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

yes no

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

Since $f(x)=\frac {x^2+3x+2}{x+2}$ is a rational function, and the limit of the denominator does not equal $0$ , you are able to apply the Quotient Law to start evaluating this limit. After that, you can use some of the other limit laws in order to finish evaluating the limit. Notice that ultimately, this limit could have been evaluated by ’plugging in’ $x=2$ into the function. In other words, if $f(x) = \frac {x^2+3x+2}{x+2}$ , then $\displaystyle \lim _{x\to 2} f(x) = f(2) = 3,$ and it turns out that $f(x)$ is a ’nice’ function at $x=2$ !

Hopefully you noticed that in the previous two examples, the given limits were relatively easy to evaluate by simply ’plugging in’ an $x$ value. In fact, one of the first things you should do when trying to evaluate a limit like $\displaystyle \lim _{x \to c} f(x)$ is to check whether plugging in $x=c$ into $f(x)$ will give you a defined number. If $f(c)$ is defined, then you can conclude that $\displaystyle \lim _{x \to c} f(x) = f(c)$ and the limit has been evaluated! Of course, this will not always be the case: the limit of a function at $x=c$ is only equal to the function value at $x=c$ if $f(x)$ is a ’nice’ function at $x=c$ , as we discussed previously. So how do you determine if a function is ’nice’ at $x=c$ or not? Good question! We’ll get to that soon enough...

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

yes no

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

Many of the upcoming examples involve trigonometric functions and re-writing trigonometric functions like $\tan (x), \csc (x), \sec (x),$ and $\cot (x)$ in terms of $\sin (x)$ and $\cos (x)$ . If you would like to review these topics before moving forward, take a look at this link or this link.

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

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 $\displaystyle \lim _{x\to 0}x\sin (1/x)= \lim _{x\to 0} x \cdot \lim _{x\to 0} \sin (1/x).$ We are only allowed to use this law if both limits exist, so we must check this first. Using the fact that the function $x$ is a ’nice’ function, we can reason that $\displaystyle \lim _{x\to 0}x=0.$ 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, and the original limit cannot be directly computed by limit laws.

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

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

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

yes no

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

Notice that $\displaystyle \lim _{x\to 1} \sec ^2(\sqrt {x}-1) = \lim _{x\to 1} \frac {1}{\cos ^2(\sqrt {x}-1)}.$ If we are trying to use Limit Laws to compute this limit, we would now have to use the Quotient Law to say that $\displaystyle \lim _{x\to 1} \frac {1}{\cos ^2(\sqrt {x}-1)} = \frac { \lim _{x\to 1}1}{ \lim _{x\to 1}\cos ^2(\sqrt {x}-1)}.$ 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: $\begin{align*} \displaystyle\lim_{x\to 1}\cos^2(\sqrt{x}-1) &= \cos^2(\sqrt{1}-1)\\ &=\cos^2(1-1)\\ &= \cos^2(0)\\ &=1 \end{align*}$ 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: $\lim _{x\to 1}1=1$ Hence $\frac { \lim _{x\to 1}1}{ \lim _{x\to 1}\cos ^2(\sqrt {x}-1)} = \frac {1}{1}=1$

Can this limit be directly computed by limit laws? $\displaystyle \lim _{x\to 4}{\left (\frac {2x}{x-4}-\frac {8}{x-4}\right )}$

yes no

If we are trying to use limit laws to compute this limit, we would have to use the Difference Law to say that $\displaystyle \lim _{x\to 4}\left (\frac {2x}{x-4} - \frac {8}{x-4}\right )= \displaystyle \lim _{x\to 4}\frac {2x}{x-4} - \lim _{x\to 4}\frac {8}{x-4}.$ We are only allowed to use this law if both limits exist. Let’s check each limit separately. $\begin{align*} \displaystyle\lim_{x\to 4}\frac{2x}{x-4}=\frac{\lim_{x\to 4}2x}{\lim_{x\to 4}(x-4)} \end{align*}$

and

$\begin{align*} \displaystyle\lim_{x\to 4}\frac{8}{x-4}=\displaystyle\frac{\lim_{x\to 4}8}{\lim_{x\to 4}(x-4)} \end{align*}$

We are only allowed to use this law if both limits exist. The limits in the numerators definitely exist. However, the denominator is equal to $0$ for both limits. Therefore, we cannot use the limit laws.

Even though the Quotient Law can’t be used, we can use algebra to subtract the two fractions before taking the limit to see if that helps. Indeed, this does the trick:

$\begin{align*} \displaystyle\lim_{x\to 4}\frac{2x-8}{x-4}&=\lim_{x\to 4}\frac{2(x-4)}{x-4}\\ &=\lim_{x\to 4}2\\ &=2 \end{align*}$

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

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 $\displaystyle \lim _{x\to 0} \cot (x)\sin (x) =\lim _{x\to 0} \cot (x) \cdot \lim _{x\to 0}\sin (x).$ We are only allowed to use this law if both limits exist. We know $\displaystyle \lim _{x\to 0} \sin (x) = 0$ , but what about $\displaystyle \lim _{x\to 0}\cot (x)$ ? We do not know how to find $\lim _{x\to 0}\cot (x)$ using limit laws because $0$ is not in the domain of $\cot (x)$ .

However, see what you can do with trigonometric identities. Try rewriting $\cot (x)$ as $\frac {\cos (x)}{\sin (x)}$ in order to evaluate this limit.

$\displaystyle \lim _{x\to 0} \cot (x)\sin (x) = \answer {1}$

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

yes no

Compute: $\displaystyle \lim _{x\to 0} \frac {\sin (x)}{1+x}\begin {prompt} =\answer {0}\end {prompt}$

If we are trying to use limit laws to compute this limit, we would have to use the Quotient Law to say that $\displaystyle \lim _{x\to 0} \frac {\sin (x)}{1+x} = \displaystyle \frac {\lim _{x\to 0}\sin (x)}{\lim _{x\to 0}(1+x)}.$ 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

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: $\displaystyle \frac {\lim _{x\to 0}\sin (x)}{\lim _{x\to 0}(1+x)}=\frac {0}{1}=0.$

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

yes no

The function $(1+x)^{1/x}$ is a power function with an exponent of $1/x$ . We do not have any limit laws for power functions with exponents that aren’t constant, so we cannot compute this limit using the limit laws.