# Lim x->infinity

find the values for A,B such that

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

what i thought was

t=1/x

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

Limt->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

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LCKurtz
Homework Helper
Gold Member
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.

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

LCKurtz
Homework Helper
Gold Member
$$\ln{(A-B)} = \ln{\left (\frac A B\right)}$$

$$\ln{(A-B)} = \ln{\left (\frac A B\right)}$$
No no,
log(a) - log(b) = log(a/b)

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

I approached the question by first factoring out x1/3 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 x4/3 in the numerator to 1/(1/x4/3) so i would have as the denominator f(x)/x4/3
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.