Two and Three Variable Limit Questions Find the following limits, if they exist.
$$\lim_{x,y\rightarrow 0,0}\frac{x^2 + \sin^2 y}{\sqrt{x^2+y^2}}$$
I believe we're suppose to use the squeeze theorem on this first one above. Possibly utilizing the fact that sin(y) is always between -1 and 1? I know the end result is suppose to be zero, but I'm having a hard time getting there.
$$\lim_{x,y,z\rightarrow 0,0,0}\frac{x^2 yz}{x^8 + y^4 + z^2}$$
I know the end result is suppose to be a DNE. I attempted to set multiple variables equal to zero to see what it would come out to, and got differing values. So, when $y,z = 0$, the limit $= 0$, when $y,z = x$, the limit $= \infty$. Since these values are different -> DNE. However I'm quite sure I'm not getting the full picture here either.
If someone could go over the process and logic associated with these problems, I'd greatly appreciate it. This is NOT homework, this is test prep.
 A: (1) For an application of the squeeze theorem, note that for $y \in [-\pi/2,\pi/2]$
$$\frac{2|y|}{\pi} \leq |\sin y| \leq |y|$$
and
$$(4/\pi^2)\sqrt{x^2+y^2} \leq\frac{x^2 +(4/\pi^2)y^2}{\sqrt{x^2+y^2}} \leq\frac{x^2 +\sin^2 y}{\sqrt{x^2+y^2}}   \leq \frac{x^2 +y^2}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2}.$$
Since $\lim_{(x,y) \rightarrow (0,0)}\sqrt{x^2+y^2}=0,$ the squeeze theorem implies that your limit is $0$.
(2) For the second limit, consider a path where $y = x^2$ and $z= x^4$, then
$$\lim_{(x,y,z)\rightarrow (0,0,0), y = x^2, z=x^4}\frac{x^2 yz}{x^8 + y^4 + z^2}=\lim_{x\rightarrow 0}\frac{x^8}{3x^8}=\frac1{3}\neq 0.$$
Next consider a path where $y = x$ and $z= x$, then
$$\lim_{(x,y,z)\rightarrow (0,0,0), y = z=x}\frac{x^2 yz}{x^8 + y^4 + z^2}=\lim_{x\rightarrow 0}\frac{x^4}{x^8+x^4+x^2}=\lim_{x\rightarrow 0}\frac{x^2}{x^6+x^2+1}=0$$
A: Notice that $\frac { \mathbb{sin} (y) } {y} \rightarrow 1$ for $y \rightarrow 0$. Then the first limit is the same as the limit of $\frac {x^2 + y^2} {\sqrt{x^2+y^2}}$, which is obviously $0$. The second limit is as you say.
(From a practical point of view, the fact that $\frac { \mathbb{sin} (y) } {y} \rightarrow 1$ for $y \rightarrow 0$ tells you that you may replace $\mathbb{sin} (y)$ with $y$. Note,though, that this works only for $y \rightarrow 0$, a "detail" many forget...)
