How do I find a limit with two variables?

[Jarfi]

How do I find a limit with two variables?

Homework Statement

lim(x->0)((1+ax)^(1/3)/(xln(1+x))-(arctan(bx)/(x^3))

http://www.wolframalpha.com/input/?i=lim%28x-%3E0%29%28%281%2Bax%29^%281%2F3%29%2F%28xln%281%2Bx%29%29-%28arctan%28bx%29%2F%28x^3%29%29
I understand that since the denominator x^3ln(1+x)---> 0 then the counter must also go to zero.

But what I do not understand is that since the counter has TWO variables, I always get a function of those two, instead of a solution of each.

The answer in WA isn't even sensible eitherUPDATE:

I have figured some of it, found out that

a= -3/2 and b= 1

Now how do I find the limit?

 

[Dick]

Homework Statement

lim(x->0)((1+ax)^(1/3)/(xln(1+x))-(arctan(bx)/(x^3))

http://www.wolframalpha.com/input/?i=lim%28x-%3E0%29%28%281%2Bax%29^%281%2F3%29%2F%28xln%281%2Bx%29%29-%28arctan%28bx%29%2F%28x^3%29%29



I understand that since the denominator x^3ln(1+x)---> 0 then the counter must also go to zero.

But what I do not understand is that since the counter has TWO variables, I always get a function of those two, instead of a solution of each.

The answer in WA isn't even sensible either


UPDATE:

I have figured some of it, found out that

a= -3/2 and b= 1

Now how do I find the limit?

What I don't understand is what the expression you are trying to find the limit of is. Your parentheses don't match up. Is it $\frac{(1+ax)^{1/3}}{xln(1+x)-\frac{arctan(bx)}{x^3}}$? Or $\frac{(1+ax)^{1/3}}{xln(1+x)}-\frac{arctan(bx)}{x^3}$? Or something else?

 

[Jarfi]

What I don't understand is what the expression you are trying to find the limit of is. Your parentheses don't match up. Is it $\frac{(1+ax)^{1/3}}{xln(1+x)-\frac{arctan(bx)}{x^3}}$? Or $\frac{(1+ax)^{1/3}}{xln(1+x)}-\frac{arctan(bx)}{x^3}$? Or something else?

it is:

$\frac{(1+ax)^{1/3}}{xln(1+x)}-\frac{arctan(bx)}{x^3}$

And i have solved it now