If f(x,mx) is algebraically reduced to contain only 'm', does it prove the limit does not exist?

Question

Suppose you trace the path $y=mx$ of a multivariable function $f(x,y)$ to find the limit as $(x,y)\to(0,0)$. If $f(x,mx)$ is algebraically reduced to contain only '$m$', does it prove the limit does not exist?

For example,

Consider $\lim_{(x,y)\to(0,0)}f(x,y)$, where $f(x,y)=\frac{(x+y)^2}{x^2+y^2}$.

If we let $y=mx$, then we can algebraically reduce $f(x,mx)$ to $f(1,m) $ where $\lim_{(1,m)\to(0,0)}\frac{(1+m)^2}{1+m^2}$.

The idea is that since $\{m|-\infty < m < \infty\}$, you can plug in any real number and receive a different limit verifying that the limit does not exist.

This is the idea my professor is arguing.

However, I was under the impression that if by direct substitution of $\lim_{(x,y)\to(0,0)}f(x,y)$ is an indeterminate form, we should then attempt to disprove the limit exists by taking paths and finding at least two that are unequal.

I still can not seem to agree with my professor's approach. Even though intuitively it seems correct, if we need two paths how can taking only one path provide for us enough information on the limits existence?

I have independently studied the $\epsilon-\delta$ proof and would prefer using this method as opposed to determining two out of an infinite number of paths, however since I am only in Calculus III, I do not think this is the method my professor wants to me to use.

If my professor is correct, I would appreciate any help to improve my intuition. If he is not, I would appreciate if you would help me formulate an argument I can present to him.

Answer 1

You seem to have missed the point of your professor's argument. When you do the substitution $y=mx$ you are studying the function along a line of slope $m$ that passes through the origin. For this function, we discover that the value only depends on $m$. Another way to say that is to note $$f(x,y)=\frac {(x+y)^2}{x^2+y^2}=\frac {(1+\frac yx)^2}{1+(\frac yx)^2}$$ and this function that we thought depended on two variables only depends on their ratio. Along a horizontal or vertical line (using the original definition so we do not divide by $0$) the function is $1$. Along the line $m=1$ the function is $2$. That means that there are points arbitrarily close to the origin where the function is $1$ and others where the function is $2$. Therefore there cannot be a limit at the origin.

Answer 2

There are two problematic statements in your question.

if we need two paths how can taking only one path provide for us enough information on the limits existence?

But your professor has taken more than one path. In fact, he has computed the limit for infinitely many paths $y=mx$: different values of $m$ give different paths. Your calculation shows that when you choose a different $m$ and thus a different path, you get a different limit (because changing $m$ will generally change the limit you computed). Hence the limit does not exist.

I was under the impression that if by direct substitution of lim(x,y)->(0,0)[f(x,y)] is an indeterminate form, we should then attempt to disprove the limit exists

No. Just because you get an indeterminate form does not suggest the limit does not exist. It simply tells you you must use a different approach to analyzing the problem.

For example, take the limit of $\sin(x-y)/(x-y)$ as we approach the origin. Simply “plugging in” gives an indeterminate form. Nevertheless the limit does exist and equals $1$. So your “attempt to disprove the limit exists” would be in vain.