Determine whether $f$ is continuous or discontinuous from the right or left at $x_0=0$ of $f(x)= 1$ if $x=0$, $ x \sin \frac{1}{x}$, otherwise 
Determine whether $f$ is continuous or discontinuous from the right or left at $x_0=0$, given:
  $$f(x)=
\begin{cases}
1, & x=0\\
x \sin\frac{1}{x}, &  x \neq 0
\end{cases}$$ 

from the left: $\lim_{x \rightarrow 0^-} x \sin \frac{1}{x}=0 \neq f(0) =1$
from the right: $\lim_{x \rightarrow 0^+} x \sin \frac{1}{x}=0 \neq f(0) =1$
It follows, by definition, that $f$ is discontinuous from the right or left at $x_0=0$
Is my argumentation correct? sufficient? Much appreciated
 A: We know that The value of x can take a maximum of 1 and a minimum of negative one!
$$-1 \leq \sin x \leq 1$$
So the same applies to 
$$-1 \leq \sin \frac{1}{x} \leq 1$$
Now let's multiply the inequality with $x$,
$$-x \leq x\sin \frac{1}{x} \leq x$$
Taking the limit for all three quantities!
$$\lim_{\text x \rightarrow 0}-x \leq \lim_{\text x \rightarrow 0} x\sin \frac{1}{x} \leq \lim_{\text x \rightarrow 0} x $$
Know that 
$$\lim_{\text x \rightarrow 0}-x =0 $$
$$\lim_{\text x \rightarrow 0} x=0$$
So limit for $x \sin \frac{1}{x}$ is 
$$\lim_{\text x \rightarrow 0} x\sin \frac{1}{x}=0$$
Now let's check if the function is continuous
By definition of continuity we must have 
$$\lim_{\text x \rightarrow 0} x\sin \frac{1}{x}= f(0)$$
Obviously the above is false since
$$\lim_{\text x \rightarrow 0} x\sin \frac{1}{x}=0 \neq f(0)=1$$
Addendum 
The squeeze theorem sates as follows for the function exists on interval $[a,b]$
There exists a value c such that 
$$a(x) \leq b(x) \leq c(x) $$
$$\lim_{\text x \rightarrow c}a(x) =L $$
$$\lim_{\text x \rightarrow c}c(x)= L$$
We can conclude that 
$$\lim_{\text x \rightarrow c}b(x) =L $$
