Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ A: If we take $y=mx$, then $$\frac{\sin(xy^2)}{(x^4+y^4)}=\frac{\sin(m^2x^3)}{m^2x^3}\frac{m^3}{(1+m^4)\sqrt{1+m^2}}$$
Any hint* will be appreciated.
A: $$\lim_{(x,y)\to(0,0)}\frac{x^2 y^2\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}=\lim_{r\to0}\frac{r^4\cos^2(\theta)\sin^2(\theta)\sin\left(r^3\cos(\theta)\sin^2(\theta)\right)}{r^5\left(\cos^4(\theta)+\sin^4(\theta)\right)}$$If $\sin\theta\cdot\cos\theta=0$, the numerator is zero. If $\sin\theta\cdot\cos\theta\ne0$, using $\lim_{x\to0}\sin x/x=1$, we get$$\begin{align*}&=\lim_{r\to0}\frac{r^4\cos^2(\theta)\sin^2(\theta)\cdot r^3\cos(\theta)\sin^2(\theta)}{r^5\left(\cos^4(\theta)+\sin^4(\theta)\right)}\\&=\lim_{r\to0}\frac{r^2\cos^3(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}\end{align*}$$
Now $\cos^4(\theta)+\sin^4(\theta)=(\cos^2(\theta)+\sin^2(\theta))^2-2\cos^2(\theta)\sin^2(\theta)=1-\frac12\sin^2(2\theta)\ge1/2$ and thus $\frac1{\cos^4(\theta)+\sin^4(\theta)}\le2$ which means the factor $\frac{\cos^3(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}$ is bounded. Thus the limit is $0$.
A: Since $|\sin t| \leq |t|$ for all $t\in\mathbb{R}$,
$$
|x^2 y \sin(xy^2)| \leq |x|^3 |y|^3 \leq \frac{x^6+y^6}{2}\,.
$$
the second inequality by the AM-GM inequality. This means
$$
\frac{|x^2 y \sin(xy^2)|}{(x^4+y^4)\sqrt{x^2+y^2}}
\leq \frac{1}{2} \frac{(x^6+y^6)}{(x^4+y^4)\sqrt{x^2+y^2}}
\leq \frac{1}{2}\frac{2\max(x^6,y^6)}{\max(x^4,y^4)\sqrt{\max(x^2,y^2)}}
= \max(|x|,|y|)
$$
and $\lim_{(x,y)\to (0,0)} \max(|x|,|y|) = 0$, so you can conclude by the squeeze theorem.
A: Hint:
For small enough $x,y$ such that the ordered pair $(x,y)\ne (0,0)$, we have $|\sin(xy^2)|\lt |xy^2|$
$|y|=\sqrt{y^2}\le\sqrt{y^2+x^2}$ and so by similar arguments, we have: 
$\big|\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}-0\big|\le \frac{|x^3|y^2}{x^4+y^4}\le \frac{|x|x^2}{\sqrt{x^4+y^4}}\le |x|\le \sqrt{x^2+y^2}$
Can you fill in the gaps and finish using definition of limit?
A: $$\lim_{(x,y)\to(0,0)}\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}} \\
= \lim_{(x,y)\to(0,0)}\frac{(xy)^3 }{(x^4+y^4)\sqrt{x^2+y^2}}\frac{\sin(xy^2)}{xy^2}$$
Let $y=x\tan\theta$
$$\lim_{(x,y,x\tan\theta)\to(0,0,0)}\frac{(x^{{6} }\tan^3\theta)}{x^4(1+\tan^4\theta)x\sqrt{1+\tan^2\theta}}\frac{\sin(xy^2)}{xy^2}\\
=\lim_{(x,y,x\tan\theta)\to(0,0,0)}\frac{(x\tan^3\theta)}{(1+\tan^4\theta)\sqrt{1+\tan^2\theta}}\frac{\sin(xy^2)}{xy^2}$$
The denominator involving the $\tan\theta$ would never be less than 1. This implies that the above limit shrinks to "0" due to $x$ in the numerator. Hence,
$$\lim_{(x,y)\to(0,0)}\frac{x^2 y\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}=0$$
A: (I'm leaving you some work.)
First show that
$\cos^4\theta+\sin^4\theta\ge{1\over 4} \tag 1 $
Justify making the numerator $x^3y^3.$
Take the absolute value of your expression.
Now, trig substitution for $x$ and $y.$
In the numerator, replace $|\sin\theta|$ and $|\cos\theta|$ by $1$ to make it bigger.
In the denominator, use (1) to make it smaller.
Now calculate a limit as $r\to 0^+.$
