Question

Is the function f defined by

$f(x) = \begin{cases} x, & \quad \text{if } n{\leq1}\\\ 5, & \quad \text{if } n{ >1} \end{cases}$

continuous at x=0? At x=1? At x=2?

Answer 1

The given function f is $f(x) = \begin{cases} x, & \quad \text{if } n{\leq1}\\\ 5 & \quad \text{if } n{ >1} \end{cases}$
At x = 0,
It is evident that f is defined at 0 and its value at 0 is 0.
Then, $\lim\limits_{x \to 0}$ f(x) = $\lim\limits_{x \to 0}$ x = 0
∴$\lim\limits_{x \to 0}$ f(x) = f(0)

Therefore, f is continuous at x = 0
At x = 1,
f is defined at 1 and its value at 1 is 1.
The left hand limit of f at x=1 is,
∴$\lim\limits_{x \to 1^-}$ f(x) = $\lim\limits_{x \to 1^-}$x=1
The right hand limit of f at x = 1 is,
∴$\lim\limits_{x \to 1^+}$ f(x) = $\lim\limits_{x \to 1^+}$+(5) = 5
∴$\lim\limits_{x \to 1^-}$ f(x) ≠ $\lim\limits_{x \to 1^+}$f(x)

Therefore, f is not continuous at x = 1

At x = 2,
f is defined at 2 and its value at 2 is 5
Then, $\lim\limits_{x \to 2}$ f(x) = $\lim\limits_{x \to 2}$ (5) = 5
∴$\lim\limits_{x \to 2}$ f(x) = f(2)

Therefore, f is continuous at x=2