Does the limit exist for $y=\sqrt{x}\sin \frac{1}{x}$? For the graph of   $$y=\sqrt{x}\sin \frac{1}{x},$$
Do the following limits exist? If so, what is it?
(a) $\lim_{x \to 0^+} f(x)$
(b) $\lim_{x \to 0^-}f(x)$
(c) $\lim_{x \to 0}f(x)$
By the way, the graph is $\sin \frac{1}{x}$ inside of the parabola $x=y^2$
Here's what I got
(a)  Yes, the limit is $0$
(b)  No
(c) No
I am confused on (a) because if $x$ approaches $0$ from the right hand side, then, according to the parabola, it reaches the limit $0$. But if $x$ approaches $0$ from the right hand side, according to $\sin\frac{1}{x}$, there is not limit. Please help?
 A: Well, let's look at sin ($\frac{1}{x}$) first. As x approaches 0 from either direction, $x\rightarrow 0$, so $\frac{1}{x}\rightarrow +\infty $. But by the definition of sine,|sin t| $\leq 1$ where $t\in \mathbb R$. So consider $(a,b) \subset \mathbb R $ where $0\in $(a,b). Then there will be 3 possible extreme values of y= f(x):
1) $x\in (a,b)$ such that $\frac{1}{x}=2n\pi$ where $n\in \mathbb N$.  
2) $x\in (a,b)$  such that $\frac{1}{x}=2n\pi$ + $\frac{\pi}{2}$
and 
3)  $x\in (a,b)$  such that $\frac{1}{x}=2n\pi$ + $\frac{3\pi}{2}$
Therefore,  sin ($\frac{1}{x}$) oscillates back and forth between 1 and -1
in any open neighborhood of x -> 0. Technically,due to the $\sqrt x$ term, f(x) is undefined at x < 0. But since we can approach 0 from the left and f(0) is defined, we can take the absolute value of x and obtain the same result in calculating the limit.
Now consider $\sqrt x$ sin $(\frac{1}{x})$. By the Cauchy Schwartz inequality and the result above: 
| $\sqrt x$ sin $(\frac{1}{x})$| $\leq$ |$\sqrt x$||sin $(\frac{1}{x})$|$\leq$ | $\sqrt x$| < $\epsilon$ where $\epsilon$ = $\sqrt\delta$ whenever |x| < $\delta $. So $lim_{|x|\rightarrow 0}$  $\sqrt x$ sin $(\frac{1}{x})$ = 0!      
Here's the graph to justify our limit visually: 
 
