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The textbook I'm using only contains examples for very simple functions, like $$\lim_{(x,y) \rightarrow (0, 1)}{x^2 + y^2 + 2}$$

In this case, I can just break up the function into its composite parts:

$$\lim_{(x, y) \rightarrow (0, 1)}{x^2} + \lim_{(x, y) \rightarrow (0, 1)}{y^2} + \lim_{(x, y) \rightarrow (0, 1)}{2} = 0 + 1 + 2 = 3$$

For more complex functions, it has some examples for how to verify a given limit. But none for how to compute it in the first place. Suppose I am given something like one of the following problems:

$$\lim_{(x,y) \rightarrow (0,0)} \frac{e^{xy} - 1}{y}$$ $$\lim_{(x,y) \rightarrow (0,0)} \frac{\cos {(xy)} - 1}{x^2y^2}$$ $$\lim_{(x,y) \rightarrow (0,0)} \frac{xy}{x^2 + y^2 + 2}$$

How do I begin if I want to compute the limit? What is the general approach?

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Sometimes, is common try to find some paths going to the origin with different limits. For example, $x=y\to 0$, $x=0,y\to 0$, etc... if you find two paths with different limits, the limit does not exists. –  Sigur Sep 15 '12 at 2:02
    
@Sigur: So if I try a few, and get the same limit, then I use the $\epsilon$-$\delta$ method to verify that it is indeed the limit? –  Matthew Sep 15 '12 at 2:07
    
Or you can try to modify your function to simplify it. See the answer below. –  Sigur Sep 15 '12 at 2:12

1 Answer 1

I usually find it handy to use a linear change of coordinates so that we are evaluating the limit as $(x,y)\to (0,0)$, and then change to polar coordinates. In the first example, this gives $$\begin{align} \lim\limits_{(x,y)\to(0,0)}\frac{e^{xy}-1}{y} &=\lim_{r\to 0}\frac{e^{r^2\sin\theta\cos\theta}-1}{r\sin\theta}\\ &=\lim_{r\to 0}\frac{\sum\limits_{n=0}^\infty (r^2\sin\theta\cos\theta)^n-1}{r\sin\theta}\\ &=\lim_{r\to 0}\sum\limits_{n=1}^\infty r^{2n-1}\sin^{n-1}\theta\cos^n\theta\\ &=\lim_{r\to 0} r\left(\sum\limits_{n=1}^\infty r^{2n-2}\sin^{n-1}\theta\cos^n\theta\right)=0\\ \end{align}$$ using the Taylor expansion of $e^{r^2\sin\theta\cos\theta}$. This has the advantage of detecting when limits do not exist, since the limit as $(x,y)\to(0,0)$ exists iff the limit as $r\to 0$ does not depend on $\theta$.

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