jayjay5531
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Can you say that lim(x,y)-->(0,0) f(x,y) is the same as limx-->0 [limy-->0 f(x,y)]?
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Question about limits of functions of two variables?
- Thread starter jayjay5531
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Can you say that lim(x,y)-->(0,0) f(x,y) is the same as limx-->0 [limy-->0 f(x,y)]?
Short answer is no.
Can you say that lim(x,y)-->(0,0) f(x,y) is the same as limx-->0 [limy-->0 f(x,y)]?
Longer answer: Let f(x,y) be given by
f(x,y) = \(\displaystyle \frac{x^2 + y}{x + y^2}\)
\(\displaystyle \lim_{x\to 0} [\lim_{y \to 0} f(x,y)] = \lim_{x\to 0} \frac{x^2}{x} = \lim_{x\to 0} x = 0\)
Now, if that were the true limit consider x and y going to zero along the line
y = a x
where a is not zero. Then
\(\displaystyle \lim_{(x,y)\to (0,0)} f(x,y) = \lim_{(x,y)\to (0,0)} \frac{x^2 + a x}{x + a^2 x^2} =\lim_{x\to 0} \frac{x (x + a)}{x (1 + a x)} = \lim_{x\to 0} \frac{x + a}{ (1 + a x)} = a \)
and the limit would depend on how you got to (0,0). But that can't be the case so the limit is undefined.
Edit to add: If you let y go to zero first, then you are just working along a line. That is |y| < \(\displaystyle \delta_y\). After that you are just working along another line, i.e. |x| < \(\displaystyle \delta_x\). However, the limit as (x,y) goes to zero is a circle \(\displaystyle \sqrt{x^2 + y^2} \lt \delta\). Now, if the limit exists, you could let y go to zero first and then x. Right?
Last edited:
HallsofIvy
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If you have \(\displaystyle \lim_{(x,y)\to (0, 0)} f(x, y)\), then often it is simplest to change to polar coordinates. That way, how "close" to (0, 0) a point is depends entirely on "r", not "\(\displaystyle \theta\)". If you can show that the limit, as r goes to 0 is independent of \(\displaystyle \theta\), then that will be the limit "as (x, y) goes to (0, 0)".
In the example Ishuda gave, \(\displaystyle \frac{x^2+ y}{x+ y^2}\), in polar coordinates, that is \(\displaystyle \frac{r^2cos^2(\theta)+ rsin(\theta)}{rcos(\theta)+ r^2sin^2(\theta)}= \frac{rcos^2(\theta)+ sin(\theta)}{cos(\theta)+ r sin^2(\theta)}\), for \(\displaystyle r\ne 0\), so that the limit, as r goes to 0, is \(\displaystyle \frac{sin(\theta)}{cos(\theta)}= tan(\theta)\), dependent on \(\displaystyle \theta\). Of course, going to (0, 0) along the line "y= ax" means the slope of the line, \(\displaystyle tan(\theta)\), is "a".
In the example Ishuda gave, \(\displaystyle \frac{x^2+ y}{x+ y^2}\), in polar coordinates, that is \(\displaystyle \frac{r^2cos^2(\theta)+ rsin(\theta)}{rcos(\theta)+ r^2sin^2(\theta)}= \frac{rcos^2(\theta)+ sin(\theta)}{cos(\theta)+ r sin^2(\theta)}\), for \(\displaystyle r\ne 0\), so that the limit, as r goes to 0, is \(\displaystyle \frac{sin(\theta)}{cos(\theta)}= tan(\theta)\), dependent on \(\displaystyle \theta\). Of course, going to (0, 0) along the line "y= ax" means the slope of the line, \(\displaystyle tan(\theta)\), is "a".