# Showing that a limit exists in $\mathbb{R}^n$

In calculus, I learned that there's one way to show that a limit for $f(x,y)$, say, exists at $(x_0,y_0)$ and another way to show that it does not exist. Specifically, if I suspect that the limit does not exist, I might choose a few different paths toward $(x_0,y_0)$, find their respective limits and hope that each evaluates to something different. If I think it does exist, I figure out, one way or another, what the limit is and then show that for all $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x,y)-L| < \epsilon$ if $\sqrt{(x-x_0)^2 + (y-y_0)^2} < \delta$. We can't use the first method here because we need to check every path into $(x_0,y_0)$.

However, I'm wondering what "every path" means exactly. For example, using the $\delta$-$\epsilon$ definition of the limit,

$$\lim_{(x,y)\rightarrow(0,0)} \frac{5x^2y}{x^2+y^2} = 0.$$

But could I instead evaluate the limit when $y=0$ and when $y=kx$? — that is, show that

$$\lim_{(x,0)\rightarrow(0,0)} \frac{5x^2y}{x^2+y^2} = \lim_{(x,0)\rightarrow(0,0)} \frac{0}{x^2 + 0} = 0$$

and that

$$\lim_{(x,kx)\rightarrow(0,0)} \frac{5x^2y}{x^2+y^2} = \lim_{(x,kx)\rightarrow(0,0)} \frac{5x^2(kx)}{x^2 + (kx)^2} = \lim_{(x,kx)\rightarrow(0,0)} \frac{5k}{k^2 + 1}x = 0$$

(Or for that matter, find the limits when $y=kx$ and when $x=ky$.) Would such a technique show that the limit tends to $0$ along every path? These are all, of course, straight paths; some paths aren't straight. But in many cases (whatever that means), paths will be pretty straight if we zoom in enough. (To be clear, I know these are not math terms :).) This wouldn't work for a wiggly path, like $y = \sin(1/x)$, but then I'm not even sure that's a valid path to take: Wouldn't such a path have to approach $0$ as well?

Essentially, I guess my real question is: How does one check every path in evaluating a limit? Or is the whole point that we can't? — instead we have to show the contrapositive: If there are two limits that converge to different values, then the limit does not exist. If it's impossible, what topic would I read up on to learn why the all-more-or-less-straight-paths approach I describe doesn't work?

Thank you! (And if I made any errors in notation for the specific paths — e.g., the bits like $(x,kx)$ — please let me know.)

• $$f(x,y) = \begin{cases} 0 &, y = 0\\\frac{x\sqrt{x^2+y^2}}{y} &, y \neq 0\end{cases}$$ has limit $0$ along every straight line, but is unbounded in every neighbourhood of $(0,0)$. – Daniel Fischer Aug 21 '13 at 16:20
• nice!An answer maybe? – Avitus Aug 21 '13 at 16:45

Computing the limit when the path approaching the point $(0,0)$ is a line of the form $y=kx$ is not enough to prove that the limit exists. Using polar coordinates your limit is

$$\lim_{\rho\rightarrow 0}\frac{5\rho^3\cos^2\theta\sin\theta}{\rho^2}=\lim_{\rho\rightarrow 0}\rho\cos^2\theta\sin\theta.$$

Now, the use of polar coordinates is not a choice of path approaching $(0,0)$: it is a convenient parametrization (or better, a choice of coordinates) for some limits of functions in $\mathbb R^2$. In the above, you are supposed to compute the limit for $\rho\rightarrow 0$ of the product

$$\rho\cdot f(\theta),$$

with $f(\theta)$ bounded function of $\theta$.. This implies that your limit converges to $0$, independently of the path chosen to approach $(0,0)$.

Well, you need to consider paths that are not necessarily along a straight line. For example you should consider also the path $(x,y) = (t,t^2)$ as $t \to 0$, and many more...

In order to prove that the limit is 0 you go with the definition. Fix $\epsilon>0$, and set $\delta=\sqrt{\epsilon/5}$. Suppose that $\sqrt{x^2+y^2}<\delta$. Then $|y| < \epsilon/5$. Now, if $\sqrt{x^2+y^2}<\delta$, then $|\frac{5x^2y}{x^2+y^2}| \leq 5|y|\frac{x^2}{x^2+y^2} \leq \epsilon \cdot 1$, as required.