Question

Find the points of discontinuity of f, where
\(f(x)=\left\{\begin{matrix} \frac{sin\,x}{x} &if\,x<0 \\   x+1& if\,x\geq0  \end{matrix}\right.\)

Answer 1

\(f(x)=\left\{\begin{matrix} \frac{sin\,x}{x} &if\,x<0 \\   x+1& if\,x\geq0  \end{matrix}\right.\)
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)=\(\frac{sin\,c}{c}\) and \(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\)(\(\frac{sin\,x}{x}\))=\(\frac{sin\,c}{c}\)
∴\(\lim_{x\rightarrow c}\) f(x)=f(c)
Therefore,f is continuous at all points x, such that x<0

Case II:
If c>0,then f(c)=c+1 and \(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\)(x+1)=c+1
∴\(\lim_{x\rightarrow c}\) f(x)=f(c)
Therefore, f is continuous at all points x, such that x>0

Case III:
If c=0,then f(c)=f(0)=0+1=1
The left-hand limit of f at x=0 is,
\(\lim_{x\rightarrow 0^-}\)f(x)=\(\lim_{x\rightarrow 0^-}\) \(\frac{sin\,x}{x}\)=1
The right-hand limit of f at x=0 is,

\(\lim_{x\rightarrow 0^+}\)f(x)=\(\lim_{x\rightarrow 0^+}\) (x+1)=1
∴\(\lim_{x\rightarrow 0^-}\) f(x)=\(\lim_{x\rightarrow 0^+}\) f(x)=f(0)
Therefore,f is continuous at x=0 From the above observations, it can be concluded that f is continuous at all points of the real line. 
Thus,f has no point of discontinuity.