how to find existence and value of limit in multivariable calculus

Question

I was in maths class and i found a question interesting. Find the limit of $\lim_{(x,y)\to (0,0)} \frac{2x}{x^2+x+y^2}$ if it exist.one of my friend did this question by transforming into polar coordinates and limit become $$\lim_{r\to 0}\frac{2\cos \theta}{r^2+\cos\theta}=\frac{2\cos \theta}{\cos\theta}=2$$ and argue the result that by the method above we sweeps out entire plane so we can do like this.But i did like this.I chose family of path given by $y^2=mx$ and put it in limit,then I got given limit as $\lim_{x\to0}\frac{2}{x+m+1}=\frac{2}{m+1}$ which depends on m which means limit doesn't exist.What is correct??Is there any general method to find existence and value of limit.I know $\epsilon-\delta$ method but for that we must know the limit.Therefore how to find limit if we don't know that.Is polar conversion helps with this problem?If one find limit as existing from any method,how can we sure that we will find limit existing and unique in other method also?

Answer 1

Your computation is right. If you substitute $y^2=mx$ you get that the limit depends on the path you take, hence it does not exist. Another way to show that it does not exist is taking the iterated limits:

$$\lim_{x\to 0}\lim_{y\to 0}\frac{2x}{x^2+x+y^2}=2$$ while

$$\lim_{y\to 0}\lim_{x\to 0}\frac{2x}{x^2+x+y^2}=0$$

so it does not exist.

The error in your argument with polar coordinates is that the limit is not well-defined for all angles, namely $\theta=\pi/2+\pi k$, $k\in \mathbb{Z}$.

A standard method to see if there is a limit is to do exactly what you did and make sure it holds for any arbitrary $r>0$ and angle $\theta \in [0,2\pi]$.

Also, trying to upperbound: Fix $\varepsilon>0$ and find $\delta$ such that for all $(x,y)$ with $\|(x,y)\| = \sqrt{x^2+ y^2} < \delta$ then

$$\left|\frac{2x}{x^2+x+y^2}\right| < \varepsilon$$ but the latter one might be tricky.

Answer 2

There is no limit. Choose $x>0, m>0$, and $y=\sqrt{mx}$, then ${2x \over x^2 +x + y^2} = {2x \over x^2+(1+m)x} = { 2 \over x+(1+m)}$, and $\lim_ {x \downarrow 0} { 2 \over x+(1+m)} = { 2 \over 1+m}$. Since $m>0$ was arbitrary, there is no limit of the original function.

Your friend's technique would need to allow $\theta$ to be arbitrary as well, otherwise the limit is only taken over radial lines.