Thread: Finding limit of this function, using Limit rules

1. Finding limit of this function, using Limit rules

I have some problems where I need to determine if the limits exist using the limit definition or limit laws. For the following 2 I think i can use the laws but would like to make sure:

$lim
](x,y)-> (0,0) = y/x^2$

I used the product rule splitting into 2 ie

$f(x,y) = y$ and $g(x,y) = 1/x^2$

This would give a limit of 0 times infinity = 0 respectively right?

Then the 2nd problem I thought I could handle the same way ie:

$lim
(x,y)-> (1,0) y/x,$

I thought i could solve it the same way, ie use the product rule with 2 equations f(x,y)=y and g(x,y) = 1/x giving

$lim
(x,y) -> (1,0) f(x,y) g(x,y) = (0)(1) = 0.$

I still need to do the proofs now but would like to know if i''m on the right track in terms of at least getting the limits right?

many thanks!

2. Originally Posted by iva
I have some problems where I need to determine if the limits exist using the limit definition or limit laws. For the following 2 I think i can use the laws but would like to make sure:

$lim
](x,y)-> (0,0) = y/x^2$

Make y approach zero along the line $y=x$ and then it follows

at once that the limit doesn't exist.

Tonio

I used the product rule splitting into 2 ie

$f(x,y) = y$ and $g(x,y) = 1/x^2$

This would give a limit of 0 times infinity = 0 respectively right?

Then the 2nd problem I thought I could handle the same way ie:

$lim
(x,y)-> (1,0) y/x,$

I thought i could solve it the same way, ie use the product rule with 2 equations f(x,y)=y and g(x,y) = 1/x giving

$lim
(x,y) -> (1,0) f(x,y) g(x,y) = (0)(1) = 0.$

I still need to do the proofs now but would like to know if i''m on the right track in terms of at least getting the limits right?

many thanks!
.

3. I'm sorry, I don't understand. For the line y=x, as y approaches zero the limit is tending to zero isn't it?

4. Originally Posted by iva
I'm sorry, I don't understand. For the line y=x, as y approaches zero the limit is tending to zero isn't it?

$\displaystyle{y=x\Longrightarrow \frac{y}{x^2}=\frac{x}{x^2}=\frac{1}{x}\xrightarro w [x\to 0^+]{}\infty$

Tonio