Discuss the continuity of the function f, where f is defined by

Solution:

The given function is $f(x)=\left\{\begin{array}{l}3, \text { if } 0 \leq x \leq 1 \\ 4, \text { if } 1

The given function is defined at all points of the interval [0, 10].

Let c be a point in the interval [0, 10].

Case I:

If $0 \leq c<1$, then $f(c)=3$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(3)=3$

$\therefore \lim _{x \rightarrow \infty} f(x)=f(c)$

Therefore, f is continuous in the interval [0, 1).

Case II:

If $c=1$, then $f(3)=3$

The left hand limit of f at x = 1 is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(3)=3$

The right hand limit of f at x = 1 is,

$\lim _{f} f(x)=\lim (4)=4$

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

If $1

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all points of the interval (1, 3).

Case IV:

If $c=3$, then $f(c)=5$

The left hand limit of f at x = 3 is,

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(4)=4$

The right hand limit of f at x = 3 is,

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(5)=5$

It is observed that the left and right hand limits of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

Case V:

If $3

$\lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all points of the interval (3, 10].

Hence, f is not continuous at x = 1 and x = 3