I am learning to solve limit with multi-variables by myself.

but I don't understand some concepts when we apply the technique.

Example (1) : $f(x,y)=\frac{x+y}{x^2+y^2}$ , $g(x,y,z)=\frac{xyz}{x^2+y^2+z^2}$ , $h(x,y)=\frac{xy}{x^2+y^2}$


when we solve $\lim_{(x,y)\to(0,0)}f(x,y)$ , why can we apply the following technique in case of $f(x,y)$

$\lim_{x\to0}f(x,0)\neq\lim_{y\to0}f(0,y) \Rightarrow \lim_{(x,y)\to(0,0)}f(x,y)$ doesn't exist ,

but we can't apply the technique in case of $g(x,y,z)$.


In example (1) , which function(s) can we apply polar , cylindrical and spherical coordinate system to replace the variables and please explain briefly.


$h(x,y)$ , we can consider $(x,y)\to(0,0)$ along the line $y=mx$

can we apply the same concept in case of $f(x,y)$ ??

And what kinds of the limit can apply the technique ??

Thanks for your help :)


You say you can't apply the technique in the case of $g(x,y,z)$. This is not correct. You can apply the technique, it just doesn't tell you anything. Let me be more specific. We have the theorem

Theorem. Let $f: \mathbb{R}^n \to \mathbb{R}^m$. If $\lim\limits_{x \to x_0} f(x) = L \in \mathbb{R}^m$, then for every sequence $(x_n)_n$ which converges to $x_0$ we have that $\lim\limits_{n \to \infty} f(x_n) = L.$

This means that if the limit exists then no matter how we approach the point, we will always approach the same number.

In the case of the first example approaching from $x=0$ and $y>0$ gives $+\infty$, a different result than from $x=0$ and $y < 0$ which gives $-\infty$. It follows that the limit cannot exist, since if it did we would have $\infty = -\infty$ which is false. (The theorem may be extended to cases in $\mathbb{R}$ for $\pm \infty$.)

In the second example we can happily attempt to do the same thing. Nothing stops us from trying. However, it won't do us any good. This is because in the second case the limit as $(x,y,z) \to (0,0,0)$ does exist. So approaching it in different ways will never give different answers, so the theorem above will never be able to disprove that the limit exists. The problem, of course, with this is that we don't know ahead of time that the limit exists. We try a few paths and they all turn out the same, so we begin to suspect that the limit might exist.

With experience there are some tricks to tell. For instance, when I look at the second example, I note that the total degree of the top is 3, and the total degree of the bottom is 2. In this case the function is going to behave like $x^3/x^2$ would in one dimension, so the limit is going to be zero. Now we must be careful, this is just intuition. You can prove that degree arguments like this work under certain restrictions, but this is a topic for a more advanced analysis course.

To prove the limit does exist, you have to use the $\epsilon-\delta$ definition of the limit. That is, for every $\epsilon >0$ you must find $\delta > 0$ such that when $|(x,y,z)-(0,0,0)|<\delta$ you have $|g(x,y,z)| < \epsilon$. One easy trick is to note that $|x| = \sqrt{x^2} \leq \sqrt{x^2 + y^2 + z^2}$ and similarly for $y$ and $z$. So $$ |g(x,y,z)| \leq \frac{(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2} = \sqrt{x^2+y^2+z^2} = |(x,y,z)|. $$ Hence if we took $\delta = \epsilon$ the previous line would show $|g(x,y,z)|\leq|(x,y,z)|< \delta = \epsilon$ whenever $|(x,y,z)|<\delta$. It follows that the limit does exist and is zero.

In the third example the degree of the top and bottom are equal so we suspect that the limit may not exist. Indeed, taking a path $x=0$ and $y \to 0$ gives 0 along the whole path, but taking a path $x=y \to 0$ gives $\frac{1}{2}$ along the whole path, so the limit does not exist.

I hope this clears things up for you.


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