# 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.

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

what is the power rule
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one