Find A, B with Limx->infinity Equation

[Dell]

find the values for A,B such that

Lim_x->infinity x*((x³+x²+ax)^(1/3)-(x³-bx)^(1/3)) = 3

what i thought was

t=1/x

Lim_t->0 1/t*((1/t³+1/t²+a/t)^(1/3)-(1/t³-b/t)^(1/3)) = 3

Lim_t->0 1/t*((1/t³+1/t²+a/t)^(1/3)-(1/t³-b/t)^(1/3)) = 3

since here we have 0/0 i can use l'hopital's law, but it looks like its going to get really ugly whith too manu terms,

also how can i solve for both A and B when i have only one equation

 

[LCKurtz]

Before you let x --> oo, take the natural log of both sides. I think a little algebra will take you farther than L'Hospital's rule. At first glance, I would say you will have a hard time finding any value of a and b that work. But perhaps "none" is the answer.

 

[Dell]

what will logs help, i have subtraction between the 2 roots

 

[LCKurtz]

$$\ln{(A-B)} = \ln{\left (\frac A B\right)}$$

 

[Bohrok]

$$\ln{(A-B)} = \ln{\left (\frac A B\right)}$$

No no,
log(a) - log(b) = log(^a/_b)

 

[LCKurtz]

Aaargh! Forgot to engage brain. In a danger zone. Still reeling from Super Bowl sunday. What can I say?

 

[andylu224]

I approached the question by first factoring out x^1/3 from the bracketted expression. with the remaining surdic expression, i rationalised the numerator in terms of cube roots so the numerator i would have
[(x² + x + a) - (x² - b)] = x + a + b in it.
I rearranged the x^4/3 in the numerator to 1/(1/x^4/3) so i would have as the denominator f(x)/x^4/3
where f(x) was the expression i used to rationalise the numerator.

using lim (f(x)/g(x)) = lim(f(x))/lim(g(x)), i evaluated the limit for the denominator and got 3 as the result.

so that means lim (x->infinity) (x + a + b) = 9

all linear functions i know of tend towards infinity as x tends towards infinity. thus, the limit is impossible no matter what value of a or b. at least that's what i got.