<s>[INST] 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
 [/INST] 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 $0$.
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