Proving existence of a limit I try to prove that the folowing limit does not exist.
$$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}}, a,b>0, a\neq b$$
I make the assumption that this limit exist. So, I try to write the limit $$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x}{\sqrt{x^2+y^2}}$$ as a limit of the function $ \dfrac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}}$ and another function, that the limits of those functions exist, and thus from the limits function algebra, the $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x}{\sqrt{x^2+y^2}}$ exists which is a contradiction, because this limit obviously does not exist. Maybe the limit $\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{\sin (ax^2+by^2)}{ax^2+by^2}=1$ can help in someway.
Any Ideas?? Thank you
 A: Suppose we have a function $f(x,y)$ such that
$$\lim_{(x,y)\to(0,0)}|f(x,y)|=0\iff\lim_{(x,y)\to(0,0)}f(x,y)=0$$
then
$$0\le\left|\frac{xf(x,y)}{\sqrt{x^2+y^2}}\right|=\frac{|x||f(x,y)|}{\sqrt{x^2+y^2}}\le\frac{|x||f(x,y)|}{\sqrt{x^2}}=|f(x,y)|$$
and hence
$$\lim_{(x,y)\to(0,0)}\left|\frac{xf(x,y)}{\sqrt{x^2+y^2}}\right|=0\iff\lim_{(x,y)\to(0,0)}\frac{xf(x,y)}{\sqrt{x^2+y^2}}=0$$
by the squeeze theorem. Apply this with $f(x,y)=\sin{(ax^2+by^2)}$.
A: As Peter Foreman's question comment indicates, the limit does exist and is equal to $0$. One way to see this is with
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}} & = \lim_{(x,y)\to(0,0)} \frac{\sin(ax^2+by^2)(ax^2+by^2)(x)}{(ax^2+by^2)\sqrt{x^2+y^2}} \\
& = \lim_{(x,y)\to(0,0)}\left(\frac{\sin(ax^2+by^2)}{(ax^2+by^2)}\right)\left(\frac{(ax^2+by^2)(x)}{\sqrt{x^2+y^2}}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The first factor's limit is $1$, as you've already noted in your question text. For the second factor, using polar coordinates so $x = r\cos(\theta)$ and $y = r\sin(\theta)$, you get
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to(0,0)}\frac{(ax^2+by^2)(x)}{\sqrt{x^2+y^2}} & = \lim_{r \to 0}\frac{(ar^2\cos^2(\theta) + br^2\sin^2(\theta))(r\cos(\theta))}{r} \\
& = \lim_{r \to 0}r^2(a\cos^2(\theta) + b\sin^2(\theta))\cos(\theta) \\
& = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, the product of the limits of the  $2$ factors in \eqref{eq1A} is $0$, showing the limit is $0$.
A: $\sin (ax^2 + by^2) < (|a|+|b|)(x^2 + y^2)$
Let $\delta  = \max(|x|,|y|)$
$|\frac {x\sin(ax^2+by^2)}{\sqrt {x^2+y^2}}| < \frac {2(|a|+|b|)(\delta^3)}{\delta}$ 
