Limits and continuity

by freya81
Tags: continuity, limits
 Sci Advisor HW Helper P: 2,002 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
 HW Helper P: 1,422 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. ----------------- 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! ----------------- iii, and iv can be done exactly in the same way.