# 2.4 Continuity  (Page 4/16)

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Requiring that $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=f\left(a\right)$ and $\underset{x\to {b}^{-}}{\text{lim}}f\left(x\right)=f\left(b\right)$ ensures that we can trace the graph of the function from the point $\left(a,f\left(a\right)\right)$ to the point $\left(b,f\left(b\right)\right)$ without lifting the pencil. If, for example, $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)\ne f\left(a\right),$ we would need to lift our pencil to jump from $f\left(a\right)$ to the graph of the rest of the function over $\left(a,b\right].$

## Continuity on an interval

State the interval(s) over which the function $f\left(x\right)=\frac{x-1}{{x}^{2}+2x}$ is continuous.

Since $f\left(x\right)=\frac{x-1}{{x}^{2}+2x}$ is a rational function, it is continuous at every point in its domain. The domain of $f\left(x\right)$ is the set $\left(\text{−}\infty ,-2\right)\cup \left(-2,0\right)\cup \left(0,\text{+}\infty \right).$ Thus, $f\left(x\right)$ is continuous over each of the intervals $\left(\text{−}\infty ,-2\right),\left(-2,0\right),$ and $\left(0,\text{+}\infty \right).$

## Continuity over an interval

State the interval(s) over which the function $f\left(x\right)=\sqrt{4-{x}^{2}}$ is continuous.

From the limit laws, we know that $\underset{x\to a}{\text{lim}}\sqrt[]{4-{x}^{2}}=\sqrt{4-{a}^{2}}$ for all values of a in $\left(-2,2\right).$ We also know that $\underset{x\to {-2}^{+}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists and $\underset{x\to {2}^{-}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists. Therefore, $f\left(x\right)$ is continuous over the interval $\left[-2,2\right].$

State the interval(s) over which the function $f\left(x\right)=\sqrt{x+3}$ is continuous.

$\left[-3,\text{+}\infty \right)$

The [link] allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

## Composite function theorem

If $f\left(x\right)$ is continuous at L and $\underset{x\to a}{\text{lim}}g\left(x\right)=L,$ then

$\underset{x\to a}{\text{lim}}f\left(g\left(x\right)\right)=f\left(\underset{x\to a}{\text{lim}}g\left(x\right)\right)=f\left(L\right).$

Before we move on to [link] , recall that earlier, in the section on limit laws, we showed $\underset{x\to 0}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=1=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right).$ Consequently, we know that $f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at 0. In [link] we see how to combine this result with the composite function theorem.

## Limit of a composite cosine function

Evaluate $\underset{x\to \pi \text{/}2}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x-\frac{\pi }{2}\right).$

The given function is a composite of $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ and $x-\frac{\pi }{2}.$ Since $\underset{x\to \pi \text{/}2}{\text{lim}}\left(x-\frac{\pi }{2}\right)=0$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at 0, we may apply the composite function theorem. Thus,

$\underset{x\to \pi \text{/}2}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x-\frac{\pi }{2}\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\underset{x\to \pi \text{/}2}{\text{lim}}\left(x-\frac{\pi }{2}\right)\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right)=1.$

Evaluate $\underset{x\to \pi }{\text{lim}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x-\pi \right).$

0

The proof of the next theorem uses the composite function theorem as well as the continuity of $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and $g\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ at the point 0 to show that trigonometric functions are continuous over their entire domains.

## Continuity of trigonometric functions

Trigonometric functions are continuous over their entire domains.

## Proof

We begin by demonstrating that $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous at every real number. To do this, we must show that $\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x=\text{cos}\phantom{\rule{0.1em}{0ex}}a$ for all values of a .

$\begin{array}{ccccc}\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\hfill & =\underset{x\to a}{\text{lim}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\left(x-a\right)+a\right)\hfill & & & \text{rewrite}\phantom{\rule{0.2em}{0ex}}x=x-a+a\hfill \\ & =\underset{x\to a}{\text{lim}}\left(\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x-a\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x-a\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\right)\hfill & & & \text{apply the identity for the cosine of the sum of two angles}\hfill \\ & =\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\underset{x\to a}{\text{lim}}\left(x-a\right)\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(\underset{x\to a}{\text{lim}}\left(x-a\right)\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\hfill & & & \underset{x\to a}{\text{lim}}\left(x-a\right)=0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{are continuous at 0}\hfill \\ & =\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}a-\text{sin}\phantom{\rule{0.1em}{0ex}}\left(0\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}a\hfill & & & \text{evaluate cos(0) and sin(0) and simplify}\hfill \\ & =1·\text{cos}\phantom{\rule{0.1em}{0ex}}a-0·\text{sin}\phantom{\rule{0.1em}{0ex}}a=\text{cos}\phantom{\rule{0.1em}{0ex}}a.\hfill \end{array}$

The proof that $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of $\text{sin}\phantom{\rule{0.1em}{0ex}}x$ and $\text{cos}\phantom{\rule{0.1em}{0ex}}x,$ their continuity follows from the quotient limit law.

As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

## The intermediate value theorem

Functions that are continuous over intervals of the form $\left[a,b\right],$ where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem    .

Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
by using integration product formula
Roha
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
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do u mean fundamental theorem ?
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