limits and continuity


by freya81
Tags: continuity, limits
freya81
freya81 is offline
#1
Nov24-05, 03:55 AM
P: 4
1a)Do these functions have limits.If the limit exists, find it with justification, if not explain why not


i) f(x,y) =x-y/x+y
ii) f(x,y) =x-y/x+y
iii) f(x,y) =xy/|x|+|y|
iv) f(x,y) =1-√(1-x)/x+ xy+y
v) f(x,y) =yx/y^6+x

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) ={2x sin y/ x+y if y >x
sin(x+y) if y≤x
Determine the points at which f is continuous.
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Galileo
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#2
Nov24-05, 05:46 AM
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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:
[tex]f(x,y)=\frac{xy}{|x|}+|y|[/tex]
and I guess it should be:
[tex]f(x,y)=\frac{xy}{|x|+|y|}[/tex]
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 [itex]|f(x,y)| \leq g(x,y)|[/itex]. Then, according to the squeeze theorem we have f(x,y)-> 0
VietDao29
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#3
Nov24-05, 06:41 AM
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Another way is to use polar coordinate system.
So change x, and y a little:
[itex]x = r \cos \theta[/itex]
[itex]y = r \sin \theta[/itex]
When x, and y tends to 0, r tends to 0, whereas [itex]\theta[/itex] 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 [itex]\theta[/itex] is. It just depends on r.
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I'll do i as an example:
[tex]\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[/tex]
That means the limits of the function depends on [itex]\theta[/itex], 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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