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.