##### 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

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