MHB Finding a limit of multivariable function

[tmt1]

I am trying to learn all the methods of finding the limit of a multivariable function. If I have

$$\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2 + y^2}$$

I can set $y = mx$ to see if the function solely depends on $m$, in which case the limit does not exist. So I would get

$$\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2 + m^2x^2}$$

Or$$\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2(1 + m^2)}$$

Or

$$\lim_{{(x, y)}\to{(0,0)}} \frac{1}{x(1 + m^2)}$$

So in this case, the function does not depend on $m$, therefore this method is inconclusive. (Is it possible to use this method to evaluate a limit, or just to prove that it does not exist?).

I can try to use polar coordinates like this:

$$\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2 + y^2} = \lim_{{r}\to{0}} \frac{r cos\theta}{r^2 (cos^2\theta + sin^2cos\theta)} = \lim_{{r}\to{0}} \frac{cos\theta}{r (cos^2\theta + sin^2cos\theta)}$$

I'm not sure how to simplify from here, so it appears this method is inconclusive.

Or, I can try to evaluate the different limits along the x and y axes.

So the limit along the x-axis would be
$$\lim_{{x}\to{(0)}} \frac{x}{x^2} = \infty$$And the limit along the y-axis would be

$$\lim_{{y}\to{(0)}} \frac{0}{y^2} = 0$$

Therefore, the limit does not exist, as the limits on both axes are different.

Were these methods applied correctly?

 

[Greg]

I can try to use polar coordinates like this:

$$\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2 + y^2} = \lim_{{r}\to{0}} \frac{r cos\theta}{r^2 (cos^2\theta + sin^2cos\theta)} = \lim_{{r}\to{0}} \frac{cos\theta}{r (cos^2\theta + sin^2cos\theta)}$$

$y=r\sin\theta$ not $r\sin\cos\theta$.

 

[HallsofIvy]

A few points: "\lim_{(x, y)\to (a, b)} f(x, y)= L if and only if, given any \epsilon> 0 there exist \delta> 0 such that if 0< \sqrt{(x- a)^2+ (y- b)^2}< \delta then |f(x,y- L|< \epsilon" or in words: "we can make f(x,y) as close to L as we please by making (x, y) sufficiently close to (a, b)". f(x, y) getting close to L as we "approach" (a, b) along a specific path is "necessary" for the limit to be L but not "sufficient".

For (a, b)= (0, 0) and setting y= mx we are taking the limit as (x, y) approaches (0, 0) along that specific line. If, for two different values of m (i.e. along two different lines we get two different limits that means the limit of the function itself does not exist. But many Calculus texts give an example (which I do not remember right now) to show that even if the limit is the same as we approach (a, b) along all straight lines, the limit itself does not exist because we get a different result approaching (a, b) along a parabolic path.

It is, of course, impossible to check every possible path so that method only works in showing that a given limit does not exist. Since, if we convert to polar coordinates, x= r cos(\theta), y= r sin(\theta), "nearness" to (0, 0) depends only on r, not on \theta, if the limit, as r goes to 0, exists independently of [/tex]\theta[/tex], then the limit itself exists.

Here, however, as greg1313 pointed out, you have converted to polar coordinates incorrectly. \frac{x}{x^2+ y^2}= \frac{r cos(\theta)}{r^2}= \frac{1}{r}cos(\theta) and the limit of that, as r goes to 0, does not exist.