# Finding a limit of multivariable function

• MHB
• tmt1
In summary, the conversation discusses various methods for finding the limit of a multivariable function, specifically $\lim_{{(x, y)}\to{(0,0)}} \frac{x}{x^2 + y^2}$. The first method involves setting $y = mx$ to see if the limit depends on $m$, but it is inconclusive. The second method involves using polar coordinates, but it is also inconclusive due to a mistake in converting to polar coordinates. The third method involves evaluating the limit along different axes, which results in different limits and therefore shows that the limit does not exist.
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?

tmt said:
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 $r\sin\cos\theta$.

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.

## 1. What is a limit of a multivariable function?

A limit of a multivariable function is a mathematical concept that describes the behavior of a function as the input values get closer and closer to a specific point. It is a way to measure the value of a function at a specific point, even if the function is not defined at that point.

## 2. How is the limit of a multivariable function calculated?

The limit of a multivariable function is calculated by approaching the point of interest along different paths and observing the behavior of the function as the input values get closer to the point. If the function approaches the same value regardless of the path taken, then the limit exists.

## 3. What are the conditions for a multivariable function to have a limit?

In order for a multivariable function to have a limit, it must approach the same value regardless of the path taken to the point of interest. Additionally, the limit must exist along all paths approaching the point.

## 4. How is the limit of a multivariable function different from a single-variable function?

The limit of a multivariable function is different from a single-variable function because it takes into account the behavior of the function along all possible paths approaching the point of interest. In a single-variable function, the limit is only concerned with the behavior of the function along a single path.

## 5. Why is finding the limit of a multivariable function important?

Finding the limit of a multivariable function is important because it allows us to understand the behavior of a function at a specific point. This can be useful in many applications, such as optimization problems, where we need to find the maximum or minimum value of a function at a certain point.

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