Do Functions of Two Variables Have Limits and Continuity?

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1a)Do these functions have limits.If the limit exists, find it with justification, if not explain why not


i) f(x,y) =x²-y²/x²+y²
ii) f(x,y) =x³-y³/x²+y²
iii) f(x,y) =xy/|x|+|y|
iv) f(x,y) =1-√(1-x²)/x²+ xy+y²
v) f(x,y) =y³x/y^6+x²

im having problems finding the limits especially iii) and iv). i know that you can prove a limit doesn't exist by demonstration but how do you justify it if it does exist

b) Let f be a real valued function of two real variables defined by:

f(x,y) ={2x² sin y/ x²+y² if y >x
sin(x+y) if y≤x
Determine the points at which f is continuous.
 
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If you have a feeling a limit exists you can show it does by using the definition for example. Usually simpler though is the squeeze theorem.

By the way, use brackets next time, the you've written it it looks like:
f(x,y)=\frac{xy}{|x|}+|y|
and I guess it should be:
f(x,y)=\frac{xy}{|x|+|y|}
or f(x,y)=xy/(|x|+|y|)
Also, you haven't stated the point of the limit. In this case, I guess it should be (0,0)

For iii:
You probably already found out that, if the limit exists, it must be zero. So try to find a function g(x,y) that goes to zero as (x,y)->(0,0) and for which |f(x,y)| \leq g(x,y)|. Then, according to the squeeze theorem we have f(x,y)-> 0
 
Another way is to use polar coordinate system.
So change x, and y a little:
x = r \cos \theta
y = r \sin \theta
When x, and y tends to 0, r tends to 0, whereas \theta does not (Do you know why?).
When x -> 0 and y -> 0, and the limit of a function is some value, then that value does not depend on what \theta is. It just depends on r.
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I'll do i as an example:
\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2 - y ^ 2}{x ^ 2 + y ^ 2} = \lim_{r \rightarrow 0} \frac{r ^ 2 \cos ^ 2 \theta - r ^ 2 \sin ^ 2 \theta}{r ^ 2 \cos ^ 2 \theta + r ^ 2 \sin ^ 2 \theta} = \lim_{r \rightarrow 0} \frac{r ^ 2 (\cos ^ 2 \theta - \sin ^ 2 \theta)}{r ^ 2} = \lim_{r \rightarrow 0} \cos ^ 2 \theta - \sin ^ 2 \theta
That means the limits of the function depends on \theta, which means the limit does not exist!
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iii, and iv can be done exactly in the same way.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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