# 2.3 The limit laws

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• Recognize the basic limit laws.
• Use the limit laws to evaluate the limit of a function.
• Evaluate the limit of a function by factoring.
• Use the limit laws to evaluate the limit of a polynomial or rational function.
• Evaluate the limit of a function by factoring or by using conjugates.
• Evaluate the limit of a function by using the squeeze theorem.

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

## Evaluating limits with the limit laws

The first two limit laws were stated in [link] and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

## Basic limit results

For any real number a and any constant c ,

1. $\underset{x\to a}{\text{lim}}x=a$
2. $\underset{x\to a}{\text{lim}}c=c$

## Evaluating a basic limit

Evaluate each of the following limits using [link] .

1. $\underset{x\to 2}{\text{lim}}x$
2. $\underset{x\to 2}{\text{lim}}5$
1. The limit of x as x approaches a is a : $\underset{x\to 2}{\text{lim}}x=2.$
2. The limit of a constant is that constant: $\underset{x\to 2}{\text{lim}}5=5.$

We now take a look at the limit laws    , the individual properties of limits. The proofs that these laws hold are omitted here.

## Limit laws

Let $f\left(x\right)$ and $g\left(x\right)$ be defined for all $x\ne a$ over some open interval containing a . Assume that L and M are real numbers such that $\underset{x\to a}{\text{lim}}f\left(x\right)=L$ and $\underset{x\to a}{\text{lim}}g\left(x\right)=M.$ Let c be a constant. Then, each of the following statements holds:

Sum law for limits : $\underset{x\to a}{\text{lim}}\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to a}{\text{lim}}f\left(x\right)+\underset{x\to a}{\text{lim}}g\left(x\right)=L+M$

Difference law for limits : $\underset{x\to a}{\text{lim}}\left(f\left(x\right)-g\left(x\right)\right)=\underset{x\to a}{\text{lim}}f\left(x\right)-\underset{x\to a}{\text{lim}}g\left(x\right)=L-M$

Constant multiple law for limits : $\underset{x\to a}{\text{lim}}cf\left(x\right)=c·\underset{x\to a}{\text{lim}}f\left(x\right)=cL$

Product law for limits : $\underset{x\to a}{\text{lim}}\left(f\left(x\right)·g\left(x\right)\right)=\underset{x\to a}{\text{lim}}f\left(x\right)·\underset{x\to a}{\text{lim}}g\left(x\right)=L·M$

Quotient law for limits : $\underset{x\to a}{\text{lim}}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\text{lim}}f\left(x\right)}{\underset{x\to a}{\text{lim}}g\left(x\right)}=\frac{L}{M}$ for $M\ne 0$

Power law for limits : $\underset{x\to a}{\text{lim}}{\left(f\left(x\right)\right)}^{n}={\left(\underset{x\to a}{\text{lim}}f\left(x\right)\right)}^{n}={L}^{n}$ for every positive integer n .

Root law for limits : $\underset{x\to a}{\text{lim}}\sqrt[n]{f\left(x\right)}=\sqrt[n]{\underset{x\to a}{\text{lim}}f\left(x\right)}=\sqrt[n]{L}$ for all L if n is odd and for $L\ge 0$ if n is even.

We now practice applying these limit laws to evaluate a limit.

## Evaluating a limit using limit laws

Use the limit laws to evaluate $\underset{x\to -3}{\text{lim}}\left(4x+2\right).$

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

$\begin{array}{ccccc}\underset{x\to -3}{\text{lim}}\left(4x+2\right)\hfill & =\underset{x\to -3}{\text{lim}}4x+\underset{x\to -3}{\text{lim}}2\hfill & & & \text{Apply the sum law.}\hfill \\ & =4·\underset{x\to -3}{\text{lim}}x+\underset{x\to -3}{\text{lim}}2\hfill & & & \text{Apply the constant multiple law.}\hfill \\ & =4·\left(-3\right)+2=-10.\hfill & & & \text{Apply the basic limit results and simplify.}\hfill \end{array}$

## Using limit laws repeatedly

Use the limit laws to evaluate $\underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x+1}{{x}^{3}+4}.$

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

$\begin{array}{}\\ \\ \underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x+1}{{x}^{3}+4}\hfill & =\frac{\underset{x\to 2}{\text{lim}}\left(2{x}^{2}-3x+1\right)}{\underset{x\to 2}{\text{lim}}\left({x}^{3}+4\right)}\hfill & & & \text{Apply the quotient law, making sure that.}\phantom{\rule{0.2em}{0ex}}{\left(2\right)}^{3}+4\ne 0\hfill \\ & =\frac{2·\underset{x\to 2}{\text{lim}}{x}^{2}-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{\underset{x\to 2}{\text{lim}}{x}^{3}+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the sum law and constant multiple law.}\hfill \\ & =\frac{2·{\left(\underset{x\to 2}{\text{lim}}x\right)}^{2}-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{{\left(\underset{x\to 2}{\text{lim}}x\right)}^{3}+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the power law.}\hfill \\ & =\frac{2\left(4\right)-3\left(2\right)+1}{{\left(2\right)}^{3}+4}=\frac{1}{4}.\hfill & & & \text{Apply the basic limit laws and simplify.}\hfill \end{array}$

#### Questions & Answers

What is the derivative of 3x to the negative 3
-3x^-4
Mugen
that's wrong, its -9x^-3
Mugen
Or rather kind of the combination of the two: -9x^(-4) I think. :)
Csaba
-9x-4 (X raised power negative 4=
Simon
-9x-³
Jon
I have 50 rupies I spend as below Spend remain 20 30 15 15 09 06 06 00 ----- ------- 50 51 why one more
Pls Help..... if f(x) =3x+2 what is the value of x whose image is 5
f(x) = 5 = 3x +2 x = 1
x=1
Bra
can anyone teach me how to use synthetic in problem solving
Mark
x=1
Mac
can someone solve Y=2x² + 3 using first principle of differentiation
ans 2
Emmanuel
Yes ☺️ Y=2x+3 will be {2(x+h)+ 3 -(2x+3)}/h where the 2x's and 3's cancel on opening the brackets. Then from (2h/h)=2 since we have no h for the limit that tends to zero, I guess that is it....
Philip
Correct
Mohamed
thank you Philip Kotia
Rachael
Welcome
Philip
g(x)=8-4x sqrt of 3 + 2x sqrt of 8 what is the answer?
Sheila
I have no idea what these symbols mean can it be explained in English words
which symbols
John
How to solve lim x squared two=4
why constant is zero
Rate of change of a constant is zero because no change occurred
Highsaint
What is the derivative of sin(x + y)=x + y ?
_1
Abhay
How? can you please show the solution?
Frendick
neg 1?
Frendick
find x^2 + cot (xy) =0 Dy/dx
Continuos and discontinous fuctions
integrate dx/(1+x) root 1-x square
Put 1-x and u^2
Ashwini
d/dxsinh(2xsquare-6x+4)
how
odofin
how to do trinomial factoring?
Give a trinomial to factor and I'll show you how
Bruce
i dont know how to do that either. but i really wants to know know how
Bern
Do either of you have a trinomial to present that needs factoring?
Bruce
5x^2+11x+2 and 2x^2+7x-4
Samantha
Thanks, so I'll show you how to do the 1st one and then you can try to do the second one. The method I show you is a general method you can use to factor ANY factorable polynomial.
Bruce
ok
Samantha
I have to link you to a document on how to do it since this chat does not have LaTeX
Bruce
Give me a few
Bruce
Sorry to keep you waiting. Here it is: ***mathcha.io/editor/ZvZkJUdrHpVHXwhe7
Bruce
You are welcome to ask questions if you have them. Good luck
Bruce
Can I ask questions with photos ?
Lemisa
If you want
Bruce
I don't see an option for photo
Lemisa
Do you know Math?
Lemisa
You have to post it as a link
Bruce
Can you tell me how to. Post?
Lemisa
Yes, you can upload your photo on imgur and post the link here
Bruce
thanks it was helpful
Samantha
I'm glad that was able to help you. Seriously, if you have any questions, please ask. That is not easy to understand at first so I anticipate you will have questions. It would be best to convince that you understood to attempt factoring the second trinomial you posted.
Bruce
it is use in differentiation for finding the limit if x turn to zero or infinity
let see
odofin
ok
odofin
What's this exactly?
Lemisa
calculus
Sal
approximation
Alejandro
can any body teach me calc3
Sal
what is limit?   By  By By    By