Limit of 2 variable function

[slow]

Hello everyone, I am trying to verify if I have been thinking in the right way with this problem:

lim (xy-1)/(x-1) as (x,y) approaches (1,1)

I tried a substitution like this:

s = x -1
t = xy - 1 <--- this is where I am unsure if I did something wrong/forbidden or not


Now I got a new limit:

lim s/t as (s,t) approaches (0,0)


I make the substitution t= ks (k being a constant) to get:

lim s/ks as (s,ks) approaches (0,0) and get that this function is depending on k and therefore the limit does not exist, the function approaches different values along different curves as the original (x,y) approaches (1,1).

In the back of the problem collection it does say that the limit does not exist aswell but there is no explanation whatsoever. I have pretty poor basics and got unsure whether I was doing the right thing and I hope someone can help me. If I am wrong, please tell me where and I would also appreciate if there is any easier way to solve this.

 

[pka]

Hello everyone, I am trying to verify if I have been thinking in the right way with this problem:
lim (xy-1)/(x-1) as (x,y) approaches (1,1)

Look at the limit along two different paths through \(\displaystyle (1,1).\)

\(\displaystyle y=x~\&~y=1\). Do you get two different limits?

 

[slow]

Look at the limit along two different paths through \(\displaystyle (1,1).\)

\(\displaystyle y=x~\&~y=1\). Do you get two different limits?

I tried what you said and got

For y = x the limit is 2

For y = 1 the limit is 1

So I can conclude that it is non existing. Thanks! Though I want to know how you knew which paths to pick? Also, was my method in my first post ok or wrong?

 

[pka]

I tried what you said and got
For y = x the limit is 2
For y = 1 the limit is 1
So I can conclude that it is non existing. Thanks! Though I want to know how you knew which paths to pick? Also, was my method in my first post ok or wrong?

I have do hundreds of these.

 

[slow]

Alright, it takes experience I suppose. I was looking for a general way of solving limits if there existed one.

Can anyone tell me if the way I tried to solve the problem in my first post is correct?

 

[slow]

EDIT This is really in response to the immediately previous post. Pka has already given the answer to the original post.

Thank you JeffM, for your response, much appreciated. I actually meant that if there are constants k after simplifying, there exists no limit. This is merely something I found on the net though, namely through youtubing. If you take a look around the time 31:10 in this video http://www.youtube.com/watch?v=4-liutK41v0 you will see the teacher there talking about a "test" where he tries different lines by substituting y for alpha multiplied by x, alpha being k in my case. Maybe that explains better how I was thinking. Apologies if I am unclear and I am sure you can tell that I am not very good at this.

I was substituting t for ks in order to see if any k would still be there after simplifying. From what I understood from the video, it would mean different lines or curves result in different limits, like trying an arbitrary choice of line/curve and getting different answers for each one. Is this still wrong thinking?

 

[slow]

I apologize for my stupidity. Unlike pka and many others here, my academic training was in history, not math. So I too can be a bit slow at times. I got completely lost by your calling a function of x and y a constant. Let's go back to the original problem.

\(\displaystyle f(x, y) = \dfrac{xy-1}{x-1}\)

If you (a) find a substitution of variables that changes the limit point to (0, 0), (b) replace the original function with the equivalent function in the substituted variables, (c) calculate the limit of the new function by approaching along a line of constant proportionality, and (d) find that the limit along that line depends on the constant of proportionality, you have proved that no general limit exists because the limit will differ based on the direction of approach. The original poster was right, and I have been utterly wrong. I apologize abjectly.

\(\displaystyle Let\ s = x - 1\)

\(\displaystyle Let\ t = xy - 1\)

When x = 1 and y = 1, s = 0 and t = 0.

\(\displaystyle f(x, y) = g(s, t) = \dfrac{t}{s} \)

\(\displaystyle Consider\ the\ line\ s = \alpha t\)

Notice the line includes the point s = 0 and t = 0, for any value of \(\displaystyle \alpha\).

\(\displaystyle g(\alpha t, t) = j(t) = \dfrac{t}{\alpha t} = \dfrac{1}{\alpha}\)

So, the limit of j(t) as t approaches 0 is \(\displaystyle \dfrac{1}{\alpha}\), which depends on \(\displaystyle \alpha\).

So, the limit of g(s, t) as s and t approach (0, 0) along the line \(\displaystyle s = \alpha t\ \) varies according to \(\displaystyle \alpha\) and so differs by line of approach.

So, the limit of g(s, t) as s and t approach (0, 0) does not exist.

Thus, the limit of f(x, y) as x and y approach (1, 1) does not exist.

JeffM, thank you so much, this was how I was thinking but you made it look a hundred times better the way you typed it. I was confused for a while there but with your last post even my own thoughts got more clear since I was really unsure of what was happening with the limit and if I had the correct approach. Outstanding explanation, thank you again!