Question

Discuss the continuity of the function f,where f is defined by
\(\left\{\begin{matrix} -2, &if\,x\leq-1 \\   2x,&if\,-1<x\leq 1 \\   2,&if\, x>1  \end{matrix}\right.\)

Answer 1

\(\left\{\begin{matrix} -2, &if\,x\leq-1 \\   2x,&if\,-1<x\leq 1 \\   2,&if\, x>1  \end{matrix}\right.\)
   
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)=-5 
\(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\) f(-2)=-2
∴\(\lim_{x\rightarrow c}\) 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)=-2
The left-hand limit of f at x=-1 is
\(\lim_{x\rightarrow 1^-}\) f(x)=\(\lim_{x\rightarrow 1^-}\)(-2)=-2
The right-hand limit of f at x=-1 is,
\(\lim_{x\rightarrow 1^+}\) f(x)=\(\lim_{x\rightarrow 1^+}\)(2x)=2(-1)=-2
∴\(\lim_{x\rightarrow -1}\) f(x)=f(-1)
Therefore,f is continuous at x=-1

 Case(III):
If-1<c<1,then f(c)=2c and
\(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\)(2x)=2c
∴\(\lim_{x\rightarrow c}\)=f(c)
Therefore, f is continuous at all points of the interval (-1,1).

Case(IV):
If c=1,then f(c)=f(1)=2x1=2
The left-hand limit of f at x=1 is,
\(\lim_{x\rightarrow 1^-}\) f(x)=\(\lim_{x\rightarrow 1^-}\)2x)=2x1=2
The right-hand limit of f at x=1 is,
\(\lim_{x\rightarrow 1^+}\) f(x)=\(\lim_{x\rightarrow 1^+}\)(2)=2
∴\(\lim_{x\rightarrow 1}\) f(x)=f(c)
Therefore,f 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
limx→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.