Question about the thought process at multivariable limit calculation Is the following calculation correct, using the substitution method?
$$\lim_{(x,y)\to(0,0)}\frac{{x^2 - y^2}}{{2x + y}} = \lim_{(x,y)\to(0,0)}\frac{{x(1 - y/x)}}{{2 + y/x}} = \lim_{(x,y) \to(0,0)}\frac{{1 - (y/x)^2}}{{2 + y/x}} \cdot \lim_{x\to0} x,$$ where $u = y/x \to 1/2 \cdot 0 = 0$.
I'd like to know, firstly if this is correct, and after that if there is other way to solve problems like this, thank you.
 A: You have no control over $\frac y x$ so your answer is wrong.
The expression is not defined along that lime $y=-2x$ so the limit is not even well defined.
Also along the  sequence $(\frac 1  n, \frac 1 {n^{2}}-\frac 2 n)$ you get a non-zero limit.
A: Since you are asking how to solve problems like this, I add this answer.
Look carefully at the denominator: For $y=-2x$ it is not defined while for $x\neq 0$ the numerator $x^2-y^2 = x^2-(-2x)^2 = -3x^2\neq 0$.
This suggests that if you move to $(0,0)$ along a curve close t0 $y=-2x$, you may push the value of the expression $\frac{{x^2 - y^2}}{{2x + y}}$ to $+\infty$ or $-\infty$.
Now you choose curves close to $y=-2x$. Often the following trick works quite well: Add a power $x^t$ for $t>0$. With $t$ you can control how fast you get closer to $y=-2x$ while approaching $(0,0)$:
$$y=-2x+x^t \Rightarrow \frac{{x^2 - y^2}}{{2x + y}} = \frac{x^2-(-2x+x^t)^2}{x^t}\stackrel{expanding}{=} -3\frac{x^2}{x^t}-\underbrace{x^t+4x}_{\stackrel{x\to 0}{\longrightarrow}0}$$
Now, you see if you choose, for example, $t=4$, the limit for $x\to 0$ would be $-\infty$. If you choose, for example, $t=2$, the limit would be $-3$.
So, the limit you are looking for does not exist.
