Is the function f defined by
Question:

Is the function f defined by

$f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$

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

Solution:

The given function $f$ is $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x>1\end{array}\right.$

At $x=0$

It is evident that f is defined at 0 and its value at 0 is 0.

Then, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x=0$

$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$

Thereforef is continuous at x = 0

At x = 1,

is defined at 1 and its value at 1 is 1.

The left hand limit of f at x = 1 is,

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

The right hand limit of at x = 1 is,

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(5)=5$

$\therefore \lim _{x \rightarrow 1} f(x) \neq \lim _{x \rightarrow 1} f(x)$

Therefore, $f$ is not continuous at $x=1$

At $x=2$,

is defined at 2 and its value at 2 is 5.

Then, $\lim f(x)=\lim (5)=5$

$\therefore \lim _{x \rightarrow 2} f(x)=f(2)$

Therefore, f is continuous at = 2

 

 

 

 

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