If I have a function $f(x,y)$ and I need to evaluate following limit;$\displaystyle \lim_{(x,y) \rightarrow (0,0)} f(x,y)$
Could I calculate the limit of $\displaystyle \lim_{(x,x) \rightarrow (0,0)} f(x,x)$ and that be equal to$\displaystyle \lim_{(x,y) \rightarrow (0,0)} f(x,y)$? I thought since they're both approaching the same value, then they're the same, right? Or are they not equal?
Thanks
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$\begingroup$In general, no. $\displaystyle\lim_{(x,y)\to (0,0)} {y\over x}$ does not exist: $$\eqalign{ \lim_{(x,x)\to (0,0)} {x\over x} &= \lim_{x\to0}{x \over x} = 1\cr \lim_{(x,0)\to (0,0)} {0\over x} &= \lim_{x\to0}{0 \over x} = 0\cr }$$ Even if $\displaystyle\lim_{(x,rx)\to(0,0)} f(x,rx)$ exists, $\displaystyle\lim_{(x,y)\to(0,0)} f(x,y)$ might not exist, because you might have $\displaystyle\lim_{(x,rx)\to(0,0)} f(x,rx) \not= \lim_{(x,x^2)\to(0,0)} f(x,x^2)$. (I'm sure someone will chime in with an example ...)
$\endgroup$ 2 $\begingroup$When you replace $y$ by $x$, you're actually taking the limit towards $(0,0)$ by approaching along the line $y=x$. This is usually referred to as a path. You could also substitute $y=0$ and let $x \to 0$, this would be approaching along the $x$-axis (or vice versa, via the $y$-axis).
But there are more exotic paths as well, such as the parabola $y=x^2$ mentioned in Christopher Carl Heckman's answer. In fact, that are an infinite number of paths to approach from! Just choosing one, unfortunately isn't enough. Even trying many isn't, unless the limit doesn't exist.
If a limit of a function in two variables exists, then the value of the one dimensional limits you get when approaching it along a certain path, should be independent of the path. This means that if you can find two paths that give you a different limit, the limit does not exist. This is sometimes referred to as the path test.
When you have tried a few paths and you think the limit does in fact exist, then you'll need to use other methods to actually show this. That could be the formal ($\varepsilon-\delta$) definition, or algebraic manipulations etc.
Limit does not exist
A classical example of a limit that returns the same value when approached from any line, but still doesn't exist, is: $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4+y^2}$$ Approaching along the line $y=mx$ gives: $$\lim_{x \to 0} \frac{x^2mx}{x^4+m^2x^2} = \lim_{x \to 0} \frac{mx}{x^2+m^2}=0$$for any value of $m$. But when you approach along $y=x^2$, you get: $$\lim_{x \to 0} \frac{x^2x^2}{x^4+x^4} = \lim_{x \to 0} \frac{x^4}{2x^4}= \frac{1}{2}$$ This is sufficient to conclude that the limit does not exist.
Limit exists
For the limit $$\lim_{(x,y) \to (0,0)} \frac{y^3}{x^2+y^2}$$ you can try as many paths as you want, you'll always find $0$. To show that the limit is in fact $0$, you could do the following: $$\left| \frac{y^3}{x^2+y^2} \right| = \left| y \right|\left| \frac{y^2}{x^2+y^2} \right| \le \left| y \right| \to 0$$
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