Question

Find all points of discontinuity of f, where f is defined by

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

Answer 1

$f(x) = \begin{cases} x+1, & \quad \text{if } x{\geq1}\\\ x^2+1, & \quad \text{if } x \text{<1} \end{cases}$
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case I:
If c<1, then f(c) = c2+1 and $\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ f(x2+1) = c2+1
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x<1

Case (ii):
If c = 1, then f(c) = f(1) = 1+1 = 2
then the left hand limit of f at x = 1 is,
$\lim\limits_{x \to 1^-}$f(x) = $\lim\limits_{x \to 1^-}$(x2+1) = 12+1 =2
The right hand limit of f at x = 1 is,
$\lim\limits_{x \to 1^+}$ f(x) = $\lim\limits_{x \to 1^+}$(x+1) = 1+1 = 2
∴$\lim\limits_{x \to 1}$ f(x) = f(1)
Therefore,f is continuous at x = 1

** Case(iii):**
Ifc>1, then f(c) = c+1
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (x+1) = c+1
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x>1
Hence,the given function f has no point of discontinuity.