Informative line

### Definition Of Continuity

Learn definition of continuity calculus, infinite and jump discontinuity, practice one sided continuity for piecewise functions and Graph Reflection of Discontinuity.

# Continuity at a Point

A function 'f' is continuous at $$x=a$$, if

$$\lim\limits_{ x\to a^+}f(x)=\lim\limits_{ x\to a^–}f(x)=f(a)$$

$$i.e,$$  $$\lim\limits_{ x\to a}=f(a)$$   $$(\lim\limits_{ x\to a}f(x)\,\text{will exist only if }\lim\limits_{ x\to a^+}f(x)=\lim\limits_{ x\to a^–}f(x)).$$

• $$f(a)$$ is defined as an assumption in the above definition.
• If 'f' is not continuous at $$x=a$$, we say it is discontinuous at $$x=a$$.

Continuity at  $$x=a$$

$$\Rightarrow\,f(a)=f(a^+)=f(a^-)$$

(No break at $$x=a$$)  #### A function 'f' is continuous at  $$x=3$$, if  $$\lim\limits_{ x\to 3}f(x)=7$$, then the value of $$f(3)$$ is

A –9

B 11

C 17

D 7

×

If 'f' is continuous at $$x=3$$, then  $$\lim\limits_{ x\to 3}f(x)=f(3)$$ (by definition).

$$\therefore\,f(3) =7$$

### A function 'f' is continuous at  $$x=3$$, if  $$\lim\limits_{ x\to 3}f(x)=7$$, then the value of $$f(3)$$ is

A

–9

.

B

11

C

17

D

7

Option D is Correct

# Graphical Representation of Discontinuity

• If a function $$'f'$$ is discontinuous (not continuous) at $$x=a$$, then there will be a break in the graph at $$x=a$$.    #### Consider the graph of a function $$'f'$$ as shown. The value of  $$x$$ at which $$'f'$$ is discontinuous is -

A 5

B –2

C 4

D –7

×

The graph of  $$'f'$$ is broken at $$x=–2$$. We have to remove the pen as we go below $$(–2,\,0)$$. So $$x=–2$$ is a point of discontinuity.

### Consider the graph of a function $$'f'$$ as shown. The value of  $$x$$ at which $$'f'$$ is discontinuous is - A

5

.

B

–2

C

4

D

–7

Option B is Correct

# Removable Discontinuity

• If a function $$'f'$$ is discontinuous at $$x=a$$, due to the fact that limit at $$x=a$$ exists but is not equal to $$f(a)$$.

$$i.e,$$  $$\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a–}f(x)\neq f(a)$$

Then, it is called removable discontinuity.

• The name removable is given because it can be removed by redefining.

$$f(a)=\lim\limits_{x\to a}f(x)$$

$$\lim\limits_{x\to a^+}f(x)=\lim\limits_{x\to a–}f(x)\ne\,f(a)$$  • If we redefine $$f$$ at $$x=a$$ i.e $$f(a)=\lim\limits_{x\to a}f(x)$$ the dot gets filled and graph becomes continuous.

Note : we have to evaluate the limit in such problem but the language is of continuity.

#### Let   $$f(x)=\dfrac{x^2+3x–4}{x–1}$$ ,  $$'f'$$ is discontinuous at  $$x=1$$, what should be the value of $$f(1)$$ so that $$'f'$$ becomes continuous at $$x=1$$?

A –11

B 5

C 17

D –18

×

For removing discontinuity at $$x=1$$, we redefine

$$f(1)=\lim\limits_{x\to 1}\dfrac{x^2+3x–4}{x–1}$$

$$=\lim\limits_{x\to 1}\dfrac{(x-1)(x+4)}{(x–1)}$$

$$=\lim\limits_{x\to 1}(x+4)=5$$

### Let   $$f(x)=\dfrac{x^2+3x–4}{x–1}$$ ,  $$'f'$$ is discontinuous at  $$x=1$$, what should be the value of $$f(1)$$ so that $$'f'$$ becomes continuous at $$x=1$$?

A

–11

.

B

5

C

17

D

–18

Option B is Correct

# Jump Discontinuity

If the discontinuity at $$x=a$$ is due to the fact that limit doesn't exist at $$x=a,\,i.e$$

$$\lim\limits_{x\to a^+}f(x)\ne\lim\limits_{x\to a^-}f(x)$$, we say that $$'f'$$ has a jump discontinuity at $$x=a$$.

• Such a discontinuity is reflected by a jump in the graph of function at $$x=a$$. (therefore the name).
• $$f(x)=[[x]]$$ has such a discontinuity at every integer.

#### Let, $$f(x)= \begin{cases} \dfrac{|x|}{x} & if & x\neq0 \\ 1 & if & x=0 \end{cases}$$ then, which of the following is true ?

A $$f$$ has a removable discontinuity at $$x=0$$

B $$f$$ has a jump discontinuity at $$x=0$$

C $$f$$ is continuous at $$x=0$$

D $$f$$ has an infinite discontinuity at $$x=0$$

×

Consider at  $$x=0$$,

R.H.L = $$\lim\limits_{x\to 0^+}f(x)$$ (0> 0).

$$=\lim\limits_{x\to 0^+}\dfrac{|x|}{x}$$

$$=\lim\limits_{x\to 0^+}\dfrac{x}{x}$$

$$=\lim\limits_{x\to 0^+}1=1$$

L.H.L = $$\lim\limits_{x\to 0^-}f(x)$$ (0 < 0)

$$=\lim\limits_{x\to 0^-}\dfrac{|x|}{x}$$

$$=\lim\limits_{x\to 0^-}\dfrac{x}{x}=-1$$

$$\therefore$$ R.H.L $$\ne$$ L.H.L

$$\therefore$$ jump discontinuity at, $$x=0$$

### Let, $$f(x)= \begin{cases} \dfrac{|x|}{x} & if & x\neq0 \\ 1 & if & x=0 \end{cases}$$ then, which of the following is true ?

A

$$f$$ has a removable discontinuity at $$x=0$$

.

B

$$f$$ has a jump discontinuity at $$x=0$$

C

$$f$$ is continuous at $$x=0$$

D

$$f$$ has an infinite discontinuity at $$x=0$$

Option B is Correct

# Infinite Discontinuity

• If the discontinuity is due to the fact that  $$\lim\limits_{x\to a}f(x)=\infty$$ (not a finite number) then, it is called infinite discontinuity.
• $$f(x)=tan\,x$$ has an infinite discontinuity at $$x=(2n+1)\dfrac{\pi}{2}$$ where,  $$n$$ is an integer.  $$f(x)=tan\,x$$ at $$x=\dfrac{\pi}{2}$$

$$\lim\limits _{x\to{\pi}/{2}^+}tan\,x=–\infty$$  and  $$\lim\limits _{x\to{\pi}/{2}^–}tan\,x=\infty$$

At least one of two limits R.H.L or L.H.L is infinity.

#### Which of the following function has an infinite discontinuity at indicated point ?

A $$f(x)=\dfrac{1}{x+3}$$ at $$x=–3$$

B $$f(x)=\dfrac{1}{x}$$ at $$x=2$$

C $$f(x)=\dfrac{2}{x^2}$$ at $$x=2$$

D $$f(x)=\dfrac{1}{x+7}$$ at $$x=1$$

×

For option a,

$$\lim\limits_{x\to –3}\,\dfrac{1}{x+3}=\infty$$, all the other three limit exists and have finite value (direct substitution method).

$$\therefore$$ answer is (a)

For option 'b' $$\to$$ $$\lim\limits_{x\to 2}\,\dfrac{1}{x}=\dfrac{1}{2}$$

Hence, this option is incorrect.

For option c$$\to$$ $$\lim\limits_{x\to 2}\,\dfrac{2}{x^2}=\dfrac{1}{2}$$,

Hence, this option is incorrect.

For option 'd'$$\to$$ $$\lim\limits_{x\to 1}\,\dfrac{1}{x+7}=\dfrac{1}{8}$$

Hence, this option is incorrect.

### Which of the following function has an infinite discontinuity at indicated point ?

A

$$f(x)=\dfrac{1}{x+3}$$ at $$x=–3$$

.

B

$$f(x)=\dfrac{1}{x}$$ at $$x=2$$

C

$$f(x)=\dfrac{2}{x^2}$$ at $$x=2$$

D

$$f(x)=\dfrac{1}{x+7}$$ at $$x=1$$

Option A is Correct

# One Sided Continuity

• A function $$'f'$$ is said to be continuous from the right or right continuous at a value $$x=a$$ if

$$\lim\limits_{x\to a^+}f(x)=f(a)$$

i.e  R.H.L = $$f(a)$$. (L.H.L is a different value)

• Similarly, it is said to be continuous from the left or left continuous at a value $$x=a$$ if

$$\lim\limits_{x\to a^-}f(x)=f(a)$$

i.e  L.H.L = $$f(a)$$, (R.H.L is a different value)    • $$f(x)=[[x]]$$ is continuous from right at every integral value of  $$x$$

#### Which of the following functions, whose graphs are shown, have only right continuity at $$x=2$$ ?

A B C D ×

In option 'a',

$$f(2)$$ = R.H.L $$\ne$$L.H.L

$$\to$$ so, it has right continuity at $$x=2$$.

### Which of the following functions, whose graphs are shown, have only right continuity at $$x=2$$ ?

A B C D Option A is Correct

# One sided continuity for piecewise functions

• A function $$'f'$$ is said to be continuous from the right or right continuous at a value $$x=a$$ if

$$\lim\limits_{x\to a^+}f(x)=f(a)$$

i.e  R.H.L = $$f(a)$$. (L.H.L is a different value)

• Similarly, it is said to be continuous from the left or left continuous at a value $$x=a$$ if

$$\lim\limits_{x\to a^-}f(x)=f(a)$$

i.e  L.H.L = $$f(a)$$, (R.H.L is a different value)

• $$f(x)=[[x]]$$ is continuous from right at every integral value of  $$x$$    • If $$'f'$$ is a piecewise function then we can find the right and left hand limits at all point when the definition expression of f changes.
• e.g

$$f(x)= \begin{cases} 2x+5 & if & x<1\\ x^2+4 & if & x \geq1\\ \end{cases}$$

At x=1,

$$\lim\limits_{x\to 1^+}(x^2+4)=5$$= R.H.L

$$\lim\limits_{x\to 1^-}(2x+5)=7$$ = L.H.L

$$f(1)=1+4=5$$

$$\therefore\,f(1)=$$ R.H.L $$\ne$$ L.H.L

$$\therefore\,f$$ is right continuous at $$x=1$$.  #### Let $$f(x)= \begin{cases} 2+x^2 & if & x\leq1\\ 2-x & if & 1<x <2\\ x+3 & if & x\geq2 \end{cases}$$ Then choose correct option about  $$'f'$$.

A $$f$$ is right continuity at $$x=1$$

B $$f$$ is left continuity at $$x=1$$

C $$f$$ is continuous at $$x=1$$

D Nothing can be said about continuity of  $$'f'$$ at $$x=1$$

×

At $$x=1$$

R.H.L $$=\lim\limits_{x\to 1^+}f(x)=\lim\limits_{x\to 1^+}(2-x)=2-1=1$$

L.H.L$$=\lim\limits_{x\to 1^-}f(x)=\lim\limits_{x\to 1^-}(2+x^2)=2+1=3$$

$$f(1)=2+1=3$$

L.H.L$$=f(1)\Rightarrow\,$$Left continuity at $$x=1$$

### Let $$f(x)= \begin{cases} 2+x^2 & if & x\leq1\\ 2-x & if & 1<x <2\\ x+3 & if & x\geq2 \end{cases}$$ Then choose correct option about  $$'f'$$.

A

$$f$$ is right continuity at $$x=1$$

.

B

$$f$$ is left continuity at $$x=1$$

C

$$f$$ is continuous at $$x=1$$

D

Nothing can be said about continuity of  $$'f'$$ at $$x=1$$

Option B is Correct

# Graphical Difference for Various Discontinuities

Consider the graphs of functions having various types of discontinuities.

$$'f'$$ has a removable discontinuity at $$x=a$$ in the graph shown.  $$'f'$$ has infinite discontinuity at $$x=a$$ in the graph shown.  R.H.L $$\ne$$ L.H.L (R.H.L – L.H.L) = Jump.

$$f$$ has a jump discontinuity at $$x=a$$, there is a jump in the graph at $$x=a$$.  #### The graphs of four functions are given with discontinuity nature written with it. Choose incorrect option.

A B C D ×

(1) Removable discontinuity $$\Rightarrow$$ break but no jump.

(2) Jump discontinuity $$\Rightarrow$$ jump at the point. R.H.L $$\neq$$ L.H.L

(3) Infinite discontinuity $$\Rightarrow$$ Infinite jump at $$x=a$$

(At least one of R.H.L, L.H.L is infinite)

$$\therefore$$ option 'b' is incorrect.

### The graphs of four functions are given with discontinuity nature written with it. Choose incorrect option.

A B C D Option B is Correct