Question

Examine the continuity of f, where f is defined by
\(f(x)=\left\{\begin{matrix} sin\,x-cos\,x, &if\,x\neq0 \\   -1,& if\,x=0 \end{matrix}\right.\)
 

Answer 1

\(f(x)=\left\{\begin{matrix} sin\,x-cos\,x, &if\,x\neq0 \\   -1,& if\,x=0 \end{matrix}\right.\)

It is evident that f is defined at all points of the real line.
Let c be a real number.

Case I:
If c≠0,then f(c)=sin c-cos c
\(\lim_{x\rightarrow c}f(x)\)=\(\lim_{x\rightarrow c}\) (sinx-cosx)=sin c-cos c
∴\(\lim_{x\rightarrow c}\) f(x)=f(c)
Therefore,f is continuous at all points x,such that x≠0

Case II:
If c=0,then f(0)=-1 and \(\lim_{x\rightarrow 0^-}\)f(x)=\(\lim_{x\rightarrow 0}\)(sinx-cosx)=sin0-cos0=0-1=-1
 \(\lim_{x\rightarrow 0^+}\)f(x)=\(\lim_{x\rightarrow 0}\)(sinx-cosx)=sin0-cos0=0-1=-1
∴\(\lim_{x\rightarrow 0^-}\) f(x)=f(0)=\(\lim_{x\rightarrow 0^+}\)f(x)=f(0)
Therefore,f is continuous at x=0 
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus,f is a continuous function.