# 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.

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(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

$$\lim_{(x,y,z)\rightarrow (0,0,0), y = z=x}\frac{x^2 yz}{x^8 + y^4 + z^2}=\lim_{x\rightarrow 0}\frac{x^4}{x^8+x^4+x^2}=\lim_{x\rightarrow 0}\frac{x^2}{x^6+x^2+1}=0$$

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Thanks very much for explaining it in depth, exactly what I needed. – CODe Aug 7 '14 at 17:48
@CODe: You're welcome – RRL Aug 7 '14 at 17:53

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...)

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Plugging in constants for y such as y = 0 or y = x, then checking the limit does not mean the overall limit is equal to the value obtained iirc. Therefore, isn't the argument you gave for the first limit invalid/incomplete? Additionally, the second limit is most definitely not correct/complete, as my score on that question was 2/6. – CODe Aug 7 '14 at 17:22
@CODe: Before downvoting, which seriously affects someone's reputation, it's better to think twice and maybe ask first for clarifications. There is nothing "plugged"; notice that $\mathbb{sin}(y)=\frac{ \mathbb{sin}(y)}{y} y$ and now I hope that you get it. Please cancel your downvote, now. Mentally, experienced mathematicians simply replace $\mathbb{sin}(y)$ by $y$, it has the same effect. Think of writing Taylor's series for $\mathbb{sin}(y)$, simplify and finally make $y=0$, if you want, it's the same thing. – Alex M. Aug 7 '14 at 17:51
@CODe: Concerning the second limit, the OP was almost right (in particular his conclusion is correct) but the second path that he chose does not give the limit $\infty$, but $0$, which doesn't prove anything. Since he obtained $0$ once, he should have looked for paths producing a larger (or equal) power of $x$ in the denominator than in the numerator, thus producing $\infty$ (or something finite but nonzero). For instance, choose powers such that the total power of the numerator should be $8$, like in the denominator; this gives the path $x \mapsto (x,x^2,x^4)$, giving $\frac{1}{3}$. – Alex M. Aug 9 '14 at 12:25