Continuity of a Function on an Interval Example 1

Practice Questions – Continuity (Exam-Focused)

Question 1.
If
[f(x)=\dfrac{|x-2|}{x-2},; x\neq 2,][\quad] [\text{and}\quad f(2)=0,]
show that (f(x)) is continuous everywhere except at (x=2).

Step-by-Step Solution:

1. For [(x>2)]: [(|x-2|=x-2)]
   [
   f(x)=\dfrac{x-2}{x-2}=1
   ]
2. For (x<2): (|x-2|=-(x-2))
   [
   f(x)=\dfrac{-(x-2)}{x-2}=-1
   ]
3. LHL at (x=2):
   [
   \lim_{x\to 2^-}f(x)=-1
   ]
4. RHL at (x=2):
   [
   \lim_{x\to 2^+}f(x)=1
   ]
5. [(f(2)=0)]

Conclusion:
LHL ≠ RHL ≠ [(f(2))] ⇒ Discontinuous at [(x=2)]
Continuous everywhere else ✔️

Question 2.
If
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {\dfrac{{{{x}^{2}}-1}}{{x-1}},} & {x\ne 1} \\ {3,} & {x=1} \end{array}} \right.$
check continuity at [(x=1)].

Step-by-Step Solution:

1. Simplify:
   [\dfrac{x^2-1}{x-1}][=\dfrac{(x-1)(x+1)}{x-1}][=x+1] [\quad][ (x\neq 1)]
2. Limit:
   [
   \lim_{x\to 1}(x+1)=2
   ]
3. Function value:
   [
   f(1)=3
   ]

Conclusion:
[
\lim f(x)=2 \ne f(1)=3
]
⇒ Discontinuous at [(x=1)] (removable).

Question 3.
Check continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {x+5,} & {x<-3} \\ {{{x}^{2}}-9,} & {x\ge -3} \end{array}} \right.$
at [(x=-3)].

Step-by-Step Solution:

1. LHL:
   [
   \lim_{x\to -3^-}(x+5)=2
   ]
2. RHL:
   [\lim_{x\to -3^+}(x^2-9)][=9-9=0]
3. [f(-3)=(-3)^2-9=0]

Conclusion:
LHL ≠ RHL ⇒ Discontinuous at (-3) (jump).

Question 4.
Check continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {\sin x,} & {x\ne \pi } \\ {1,} & {x=\pi } \end{array}} \right.$
at [(x=\pi)].

Step-by-Step Solution:

1. Limit:
   [\lim_{x\to \pi}\sin x] [= \sin\pi] [= 0]
2. Function value:
   [
   f(\pi)=1
   ]

Conclusion:
[
\lim f(x)=0 \ne f(\pi)=1
]
⇒ Discontinuous at [(x=\pi)].

Question 5.
Check continuity of
$ \displaystyle f(x)=\left\{ {\begin{array}{*{20}{c}} {{{e}^{x}},} & {x<0} \\ {1,} & {x=0} \\ {\cos x,} & {x>0} \end{array}} \right.$
at [(x=0)].

Step-by-Step Solution:

1. LHL:
   [
   \lim_{x\to 0^-}e^x=e^0=1
   ]
2. RHL:
   [\lim_{x\to 0^+}\cos x][=\cos 0=1]
3. Function value:
   [
   f(0)=1
   ]

Conclusion:
[\text{LHL}=\text{RHL}=f(0)][=1]
⇒ Continuous at [(x=0)].