Example on Continuity

Which one of the following functions is continuous at x = 3 ?

 

(A)\(f(x) = \left\{ \begin{array}{ll} 2 & x = 3 \\ x -1 & x > 3\\ \frac{x+3}{3} & x < 3 \\ \end{array} \right. \)

 

(B) \(f(x) = \left\{ \begin{array}{ll} 4 & x = 3 \\ 8-x & x \neq 3\\ \end{array} \right. \)

 

(C) \(f(x) = \left\{ \begin{array}{ll} x+3 & x \leq 3 \\ x -4 & x > 3\\ \end{array} \right. \)

 

(D) \(f(x) = \left\{ \begin{array}{ll} \frac{1}{x^3 - 27} & x \neq 3 \\ \end{array} \right. \)

Answer

Things you need to know

For a function to be continuous at 3, the Left Hand Limit (LHL) = Right Hand Limit = \(f(3)\)

 

(a) 

\(f(3) = 2\)

LHL = \(\lim_{h \to 0} f(3 - h)\)  = \(\lim_{h \to 0} {\frac{3-h+3}{3}}\) = 2

RHL = \(\lim_{h \to 0} f(3 + h)\) = \(\lim_{h \to 0} {3 + h -1}\) = 2

∵ LHL = RHL = \(f(3)\)

∴ continuous at x=3.

(b)

\(f(3) = 4\)

LHL = \(\lim_{h \to 0} f(3 - h)\) = \(\lim_{h \to 0} {8 - (3 - h)}\)= 5

RHL = \(\lim_{h \to 0} f(3 + h)\) = \(\lim_{h \to 0} {8 - (3 + h)}\)  = 5

∵ LHL = RHL \(\neq f(3)\)

∴ Not continuous at x = 3

 

(c)

\(f(3) = 3 + 3 = 6\)

LHL = \(\lim_{h \to 0} f(3 - h)\) = \(\lim_{h \to 0} {3-h+3} = 6\)

RHL = \(\lim_{h \to 0} f(3 + h)\) = \(\lim_{h \to 0}{3 + h - 4} = -1\)

∵ \(f(3) = LHL \neq RHL\)

∴ Not continuous at x = 3.

 

(d) 

\(f(3)\) is not defined , therefore not continuous at x = 3

3Comments
SHIVAM KUMAR @shivamkumar12
6 Nov 2019 10:16 am
Nice examples , but
there are some silly error
(b) RHL = lim h→0 f(3+h) = lim h→0 8−(3+h) = 5 // 4 was there but it should be 5
(C) LHL = lim h→0 f(3−h) = lim h→0 3-h+3 = 6 // -ve sign was not there
Rohit Panwar @panwarrohit
6 Nov 2019 11:54 am
@shivam kumar ...yes there was some mistake, now it has been corrected. thanx for make us correct.
SHIVAM KUMAR @shivamkumar12
6 Nov 2019 08:23 pm
@rohit sir welcome