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.