Continuity of $f(x,y)=x^3\sin\left(\frac 1x\right)+y^2$ at $(0,0)$ Consider $$f(x,y)=  \begin{cases} x^3\sin(1/x) +  y^2\quad &  x\ne 0 \\ y^2 &\text{otherwise}\end{cases}$$ 
Prove that $f$ is continuous at $(0,0)$. 
I do not have an experience in proving continuity for functions of several variables so any hint or outline of the proof is much appreciated.
 A: Here's what we know:  For all $x$, $-1 \leq \sin{(\frac{1}{x})} \leq 1$, which means for all $x, y$ with $x > 0$:
$$-x^{3} + y^{2} \leq x^{3} \sin{(\frac{1}{x})} + y^{2} \leq x^{3} + y^{2}$$
Now both of the functions on the left and right are continuous at $(0,0)$, and taking the limit as $(x,y) \to (0,0)$ for each of them gives $0$.  But we have:
$$\lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} -x^{3} + y^{2} \leq \lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }}  x^{3} \sin{(\frac{1}{x})} \leq \lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }}  x^{3} + y^{2}$$ which implies $\lim \limits_{\substack{(x,y) \to (0,0) \\ x > 0 }} x^{3} \sin{(\frac{1}{x})} + y^{2} = 0$.  This limit holds for all possible paths where $x > 0$.
Now, if $x < 0$, then that just means $$-x^{3} + y^{2} \geq x^{3} \sin{(\frac{1}{x})} + y^{2} \geq x^{3} + y^{2}$$ but by the same argument as above, this still proves that under the paths with $x < 0$, we still have $\lim \limits_{\substack{(x,y) \to (0,0) \\ x < 0 }} x^{3} \sin{(\frac{1}{x})} + y^{2} = 0$
