2

Consider the limit:

$$\lim_{(x,y,z)\rightarrow(2,2,0)}\frac{1}{z}(x+y)$$

This limit does not exist. To prove this, I believe I need to find two directions of approach in which the limit does not agree. I can find such directions by letting $(x,y) = (2,2)$ and approaching along the positive or negative $z$ axis to find that the limit is $+\infty$ from the positive direction and $-\infty$ from the negative direction.

However, intuitively this bothers me. I skirted around an issue in my mind by saying that this function is actually the product of two other functions, $f_1(x,y,z) = \frac{1}{z}$ and $f_2(x,y,z) = (x+y)$, then I took the limit of their product. Since I know that the limit $\lim_{(x,y,z)=\rightarrow(2,2,0)}f_2$ does not cancel the same limit of $f_1$, even though $f_1$'s limit does not exist I'm "confident" that I can just treat this as a product of limits. Should I be confident? Is the explanation actually something else?

Michael Stachowsky
  • 2,943
  • The product of the limits is the limit of the product... if the limits exist! – pwerth May 24 '17 at 16:50
  • Right, the limit of $f_1$ doesn't exist, so my reasoning seems flawed. However, I'm confident that the limit itself does not exist. This is my issue at the moment :-P – Michael Stachowsky May 24 '17 at 16:51
  • 1
    Your argument that the original limit does not exist seems fine to me – pwerth May 24 '17 at 16:52
  • It doesn't to me! Why was I able to just plug in the $x$ and $y$ values for the limit and then take the limit only on the $z$ axis? Something about breaking the limit point into two separate things but not actually treating the function as two separate things is what bothers me – Michael Stachowsky May 24 '17 at 16:53
  • @MichaelStachowsky. Your original argument was indeed flawed. It is true that if $\lim_{x \to a}f_1(x)$ and $\lim_{x \to a}f_2(x)$ both exist, then so does $\lim_{x \to a}f_1(x)f_2(x)$ and its value is the product of the two previous limits. However, in your case only the $f_2$ limit it known to exist, so you cannot treat the limit as a product of limits. Note, however, that the fact that one limit exists and the other doesn't DOES NOT imply that the limit of the product doesn't exists (though in your case that happens to be true). Consider $f_1(x) = x^2$ and $f_2(x)=1/x$. – nullUser May 24 '17 at 17:13
  • @nullUser: I agree for sure. And therein lies my issue. If I can't treat it as a product of limits, then why was I able to plug in $(2,2)$ for $(x,y)$, essentially taking the limit of $f_2$, and then proceed? In other words, why didn't I need to, say, create some parametric curve along which I'm approaching the value and show it has different limits if I approach from a positive or negative $z$? – Michael Stachowsky May 24 '17 at 17:15
  • 1
    @MichaelStachowsky, you cannot just plug in $(2,2)$ for $(x,y)$. In general, you can only plug in the value when taking the limit if the function is continuous. For instance, if you just wanted to know $\lim_{(x,y,z)\to(2,2,0)}(x+y) = 4$, you can certainly just plug in $(2,2)$ for $(x,y)$ in that case, because the function $f_2(x,y,z) = x+y$ is continuous (polynomials are continuous). That part is fine, what would NOT be fine is if you tried to plug in $(x,y,z) = (2,2,0)$ for $\frac{1}{z}$. – nullUser May 24 '17 at 17:18
  • Then I wonder: although it's not true in general the the product of the limits is the limit of the products if either limit doesn't exist, isn't that only true when the two functions have behaviour that cancels in some way? Since I don't have that behaviour, can I not claim that this is a product rule application, since I know that the special case of cancelling limits clearly does not apply? – Michael Stachowsky May 24 '17 at 17:30

1 Answers1

1

What you achieve, by showing that limits along different paths are different, is to find points arbitrarily close to $(2,2,0)$ such that their distance is greater than some fix value (in this particular case the situation is even more extreme, because your path limits go to $\pm\infty$, but even if they would go to different numbers you would have this).

The above, or what you show, implies that the limit doesn't exist: because if it did, values for the function at points very close to $(2,2,0)$ would be very similar. This can be made precise using the $\varepsilon$-$\delta$ definition of limit.

Martin Argerami
  • 205,756