Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
1  
$$f(x,y) = \begin{cases} 0 &, y = 0\\\frac{x\sqrt{x^2+y^2}}{y} &, y \neq 0\end{cases}$$ 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

2 Answers 2

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 $(0,0)$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.