Continuity in Interval

The feature of continuity can be seen on a day to day basis. For instance, the human heart is beating continuously even when the person is sleeping. A continuous function is one which can be drawn on a graph paper without lifting a pen or pencil. In this article, students will learn about Continuity in an interval.

In mathematical terms, the definition of continuity is as follows.

A function f(x) is said to be continuous at a point if the following conditions are met.

  • The function at that point exists being finite.
  • The left and right-hand limit of the function is present.
  • The limit Limx→a f(x) = f(a) where is the point

The function f (x) is declared to be continuous if the above three conditions are satisfied for every point on the interval. If any of the functions are not satisfied, then the function is discontinuous at that point.

Basic Concepts on Continuity

The following are the basic concepts of continuity:

Open And Closed Intervals

An interval is called open if it doesn’t include the endpoints. It is denoted by ( ). For instance, (1, 2) means greater than 1 and less than 2. A closed interval is one which includes all the limit points. It is denoted by [ ]. For example, [2, 5] means greater than or equal to 2 and less than or equal to 5. An open interval is termed as half-open interval if it consists of one of its endpoints. It is written as ( ]. (0, 2] means greater than 0 and less than or equal to 2 and [0, 2) is greater than or equal to 0 and less than 2.

If a, b ∈ R and a < b, the following is a representation of the open and closed intervals.

  • Open interval is indicated by (a, b) = {x : a < y < b}.
  • Closed interval is indicated by [a, b] = {x : a ≤ x ≤ b}.
  • [ a, b ) = {x : a ≤ x < b} is an open interval from a to b, inclusive of a but excluding
  • ( a, b ] = { x : a < x ≤ b } is an open interval from a to b including b but excluding
  • The mandatory condition for continuity of the function f at point x = a [considering a to be finite] is that limx→a f(x) and limx→a+ f(x) should exist and be equal to f (a).
  • The function f (x) is said to be continuous:
  1. In an open interval: if it is continuous at every point in the interval
  2. In the closed interval: if
  • f(x) is continuous in (a, b)
  • Limx→a+ f (x) = f(a)
  • limx→a f(x) = f(a)

Weierstrass Approximation Theorem

If f is a continuous real-valued function on [a,b] and if any ϵ >0 is given, then there exists a polynomial p on [a,b] such that

|f(x)-P(x)|< ϵ for all x in [a,b]

In other words, a continuous function on a closed and bounded interval can be uniformly approximated on that interval by polynomials to any degree of accuracy.

Also Read

Limits Continuity and Differentiability

Differentiation

Functions

Continuity In Interval Examples

Example 1: If the function f(x)={kcosxπ2x,when xπ23, when x=π2f(x)={\{\begin{array}{rlrlrlrlrlrl}&\frac{k\cos x}{\pi-2x},\text{when }x\neq\frac\pi2\\[3pt]&3,\text{ when }x=\frac\pi2\end{array}} be continuous at x = π / 2, then find the values of k.

Solution:

f (π / 2) = 3.

Since f(x) is continuous at x = π / 2

⇒limx→π/2 (k cosx / π − 2x) = f (π / 2)

⇒k / 2 = 3

⇒k = 6

Example 2: If f (x) = f(x)={x24x+3x21, for x1 2, for x=1f(x)={\{\begin{array}{rlrlrlrlrlrl}&\frac{{x^2}-4x+3}{{x^2}-1},\text{ for }x\neq1\\[3pt]&\text{ }2,\text{ for }x=1\end{array}}, check whether the function is continuous. Justify your answer.

Solution:

f(x)={x24x+3x21, for x1 2, for x=1f(x)={\{\begin{array}{rlrlrlrlrlrl}&\frac{{x^2}-4x+3}{{x^2}-1},\text{ for }x\neq1\\[3pt]&\text{ }2,\text{ for }x=1\end{array}}

f (1) = 2, f (1+) = limx→1+ x2− 4x + 3 / [x2−1] = limx→1+(x − 3) / (x + 1) = −1

f (1−) = limx→1−x2− 4x + 3 / [x2−1] = −1 ⇒ f (1) ≠ f (1−)

Hence, the function is discontinuous at x = 1

Example 3: What is the value of f(0) in the function f(x) = [(27 − 2x)1/3 −3] / 9 − 3 (243 + 5x)1/5,

(x ≠ 0) is continuous?

Solution:

Since f (x) is continuous at x = 0, therefore

f (0) = limx→0 f(x) = limx→0 [(27 − 2x)1/3 −3] / 9 − 3 (243 + 5x)1/5, (Form 0 / 0)

= limx→0 [1 / 3] (27−2x)−2/3(−2) / [−3 / 5] (243 + 5x)−4/5 (5)

= 2

Example 4: Let f(x)={x4x4+a,x4f(x)=\left\{ \begin{matrix} \frac{x-4}{\left| x-4 \right|}+a,\,\,\,\,\,\,x4 \\ \end{matrix} \right. then f (x) is continuous at x=4

A) a = 0, b = 0       

B) a = 1, b = 1

C) a = −1, b = 1    

D) a = 1, b = −1

Solution:

We have L.H.L. 

=limx4f(x)=limh0f(4h)=limh04h44h4+α=limh0(hh+α)=a1=\underset{x\to 4}{\mathop{\lim }}\,f(x) \\ =\underset{h\to 0}{\mathop{\lim }}\,f(4-h)\\ =\underset{h\to 0}{\mathop{\lim }}\,\frac{4-h-4}{\left| 4-h-4 \right|}+\alpha \\ =\underset{h\to 0}{\mathop{\lim }}\,\left( -\frac{h}{h}+\alpha \right)\\ =a-1

R.H.L 

=limx4f(x)=limh0f(4+h)=limh04+h44+h4+b=b+1 therefore f(4)=a+b=\underset{x\to 4}{\mathop{\lim }}\,f(x)\\ =\underset{h\to 0}{\mathop{\lim }}\,f(4+h)\\ =\underset{h\to 0}{\mathop{\lim }}\,\frac{4+h-4}{\left| 4+h-4 \right|}+b=b+1 \text \ therefore \ f(4) = a+b\\

Since f(x) is continuous at x = 4, 

 limx4f(x)=f(4)=limx4+f(x) or a1=a+b=b+1 or b=1 and a=1\underset{x\to {{4}^{-}}}{\mathop{\lim }}\,f(x)=f(4)=\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,f(x) \text \ or \ a-1=a +b=b+1 \text \ or \ b=-1 \text \ and \ a=1

Example 5: Let f(x)=limnlog(2+x)x2nsinx1+x2n,f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\log (2+x)-{{x}^{2n}}\sin x}{1+{{x}^{2n}}}, then 

A)f is continuous at x=1B)limx1+f(x)=log3C)limx1+f(x)=sin1D)limx1f(x) does not existA) f \text \ is \ continuous \ at \ x=1\\ B) \underset{x\to {{1}^{+}}}{\mathop{\lim }}\, f(x)=log3\\ C) \underset{x\to {{1}^{+}}}{\mathop{\lim }}\, f(x)=-\sin 1\\ D) \underset{x\to {{1}^{-}}}{\mathop{\lim }}\, f(x) \text \ does \ not \ exist\\

Solution:

For x<1,x2n0\left| x \right|<1,{{x}^{2n}}\to 0 as nn\to \infty and for 

x>1,1/x2n0 as n, So, f(x)={log(2+x)limx2nlog(2+x)sinxx2n+1=sinx,12[log(2+x)sinx],x<1x>1x=1\left| x \right|>1,\,\,1/{{x}^{2n}}\to 0 \text \ as \ n\to \infty ,\text \ So, \ f(x)=\left\{ \begin{matrix} \log (2+x)\, \\ \underset{\to \infty }{\mathop{\lim }}\,\frac{{{x}^{-2n}}\log (2+x)-sinx}{{{x}^{-2n}}+1}=-\sin x, \\ \frac{1}{2}[log(2+x)-sinx],\, \\ \end{matrix} \right.\,\,\,\begin{matrix} \left| x \right|<1 \\ \left| x \right|>1 \\ \left| x \right|=1 \\ \end{matrix}

Thus, 

limx1+f(x)=limx1(sinx)=sin1 and limx1f(x)=limx1log(2+x)=log3.\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,(-\sin x)=-sin1 \text \ and \ \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,\,\,log(2+x) = \log 3.

Example 6: If f(x)={x1,x<0x22x,x0,f(x)=\left\{ \begin{matrix} x-1,\,\,\,\,\,\,\,x<0 \\ {{x}^{2}}-2x,\,\,x\ge 0 \\ \end{matrix} \right., then

A)f(x) is discontinuous at x=0B)f(x) is differentiable at x=0C)f(x) is nondifferentiable at x=0,2D)f(x) is continuous at x=0A) f(\left| x \right|) \text \ is \ discontinuous \ at \ x=0\\ B) f(\left| x \right|) \text \ is \ differentiable \ at \ x=0\\ C) |f(x)| \text \ is \ non-differentiable \ at \ x=0, 2\\ D) |f(x)| \text \ is \ continuous \ at \ x=0

Solution: 

f(x)={x1,x22x,x<0x0f(x)=\left\{ \begin{matrix} \left| x \right|-1,\, \\ {{\left| x \right|}^{2}}-2\left| x \right|, \\ \end{matrix} \right.\,\,\,\,\,\,\,\,\,\begin{matrix} \left| x \right|<0 \\ \left| x \right|\ge 0 \\ \end{matrix} where x0\left| x \right|\le 0 is not possible. Thus neglecting, we get 

f(x)=x22x,x0f(x)={x2+2xx<0x22xx0 Therefore f(x)={x+2x,x<0x2xx0f(\left| x \right|)=\begin{matrix} {{\left| x \right|}^{2}}-2\left| x \right|,\,\,\left| x \right|\ge 0 \\ \end{matrix}\\ f(|x)=\left\{\begin{matrix} x^{2}+2x &x<0 \\ x^{2}-2x& x\geq 0 \end{matrix}\right. \\ \text \ Therefore \ f'(\left| x \right|)=\left\{ \begin{matrix} x+2x, & x<0 \\ x-2x & x\ge 0 \\ \end{matrix} \right.

Clearly, f(x)\left| f(x) \right| is continuous at x = 0, but non-differentiable at x = 0.

f(x)={x1,x22x,x<0x0g(x)=f(x=){1x,x<0x2+2x,0x<2x22x,x2f(x)=\left\{ \begin{matrix} \left| x \right|-1,\, \\ {{\left| x \right|}^{2}}-2\left| x \right|, \\ \end{matrix} \right.\,\,\,\,\,\,\,\,\,\begin{matrix} \left| x \right|<0 \\ \left| x \right|\geq 0 \\ \end{matrix}\\ g(x)=|f(x_=)\left\{\begin{matrix} 1-x, &x<0 & \\ -x^{2}+2x,&0\leq x<2 & \\ x^{2}-2x,&x\geq 2 & \end{matrix}\right.

Clearly, f(x)\left| f(x) \right| is discontinuous at x = 0 but non-differentiable at x = 2. 

Also,

g(x)={1,x<02x+2,0<x<2.2x2,x>2f(x)g'(x)=\left\{ \begin{matrix} -1, & x<0 \\ 2x+2, & 0<x<2. \\ 2x-2, & x>2 \\ \end{matrix} \right. \left| f(x) \right| is non-differentiable at x = 0 and x = 2.