Tricky limit involving sine I'm trying to evaluate
$$\text{lim}_{(x,y) \rightarrow (0,0)} \frac{x^4 + \text{sin}^2(y^2)}{x^4+y^4}.$$
I'm pretty sure that the limit exists and is $1$; at the very least, you get that if you approach $(0,0)$ along the lines $x=0$ and $y=0$ and $x=y$.  But I can't seem to figure out how to to show that the limit is $1$.  Thanks!

Consider
$$1-{x^4+\sin^2(y^2)\over x^4+y4}={y^4-\sin^2(y^2)\over x^4+y^4}={y^4\over x^4+y^4}\left(1-{\sin^2(y^2)\over y^4} \right)$$
Now
$$\left|{y^4\over x^4+y^4}\right|\le1$$
for all $(x,y)\not=(0,0)$ and
$$\lim_{y\to0}{\sin^2(y^2)\over y^4}=\lim_{u\to0}\left({\sin u\over u}\right)^2=1$$
That should take care of things.  (Note:  the term ${y^4\over x^4+y^4}$ by itself doesn't have a limit at $(0,0)$, but it doesn't need to; all we needed was for it to be bounded.)
