Question

Determine if f is defined by
f(x)=\left\\{\begin{matrix} x^2sin\frac{1}{x}, &if\,x\neq0 \\\ 0,&if\,x=0 \end{matrix}\right.
is a continuous function?

Answer 1

f(x)=\left\\{\begin{matrix} x^2sin\frac{1}{x}, &if\,x\neq0 \\\ 0,&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)=$c^2sin\frac{1}{c}$
$\lim_{x\rightarrow c}$f(x)=$\lim_{x\rightarrow c}$ ($x^2sin\frac{1}{x}$=($\lim_{x\rightarrow c}$ x2)($\lim_{x\rightarrow c}$ sin$\frac{1}{x}$)=c2sin$\frac{1}{c}$
∴$\lim_{x\rightarrow c}$ f(x)=f(c)
Therefore,f is continuous at all points x, such that x≠0

Case II:
If c=0,then f(0)=0 and $\lim_{x\rightarrow 0^-}$f(x)=$\lim_{x\rightarrow 0^-}$($x^2sin\frac{1}{x}$)=$\lim_{x\rightarrow 0}$($x^2sin\frac{1}{x}$)
It is known that -1≤$sin\frac{1}{x}$≤1, x≠0
$\Rightarrow$-x2≤s$sin\frac{1}{x}$≤x2
\Rightarrow$$\lim_{x\rightarrow 0}(-x2)≤$\lim_{x\rightarrow 0}$($x^2sin\frac{1}{x}$)≤$\lim_{x\rightarrow 0}$x2
$\Rightarrow$0≤$\lim_{x\rightarrow 0}$($x^2sin\frac{1}{x}$)≤0
\Rightarrow$$\lim_{x\rightarrow 0}($x^2sin\frac{1}{x}$)=0
∴$\lim_{x\rightarrow 0^-}$ f(x)=0
Similarly,$\lim_{x\rightarrow 0^+}$f(x)=$\lim_{x\rightarrow 0^+}$($x^2sin\frac{1}{x}$)=$\lim_{x\rightarrow 0}$($x^2sin\frac{1}{x}$)=0
$\lim_{x\rightarrow 0^-}$ f(x)=f(0)=$\lim_{x\rightarrow 0^+}$f(x)
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.