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

Question:

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

$f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$

Solution:

The given function $f$ is $f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

If $c<-1$, then $f(c)=-2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2)=-2$

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

Therefore, $f$ is continuous at all points $x$, such that $x<-1$

Case II:

The left hand limit of at x = −1 is,

$\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1^{-}}(-2)=-2$

The right hand limit of f at = −1 is,

$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2 \times(-1)=-2$

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

Thereforef is continuous at x = −1

Case III:

If $-1$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c\therefore \lim _{x \rightarrow c} f(x)=f(c)$Thereforef is continuous at all points of the interval (−1, 1). Case IV: If$c=1$, then$f(c)=f(1)=2 \times 1=2$The left hand limit of at x = 1 is,$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(2 x)=2 \times 1=2$\ The right hand limit of f at = 1 is,$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2=2\therefore \lim _{x \rightarrow 1} f(x)=f(c)$Thereforef is continuous at x = 2 Case V: If$c>1$, then$f(c)=2$and$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2)=2\lim _{x \rightarrow c} f(x)=f(c)$Therefore,$f$is continuous at all points$x$, such that$x>1\$

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.