# Homework Help: Indeterminate limit involving (ax^2+b)^-0.5 and infinity

1. Nov 15, 2014

### throneoo

1. The problem statement, all variables and given/known data
basically I have to evaluate the limit on LHS , the above image is from wolfram alpha . (if link doesn't work , it's
limit of (x^3)*((x^2-1)^-0.5-(x^2+1)^-0.5) as x->infinity)

2. The attempt at a solution

L'hopital's rule wouldn't work, as x^3 vanishes after the 3rd derivative , but (x^2+a)^0.5 would not terminate at higher derivatives and I can't find the limit at their 1st and 2nd derivatives.

I've tried to change variables such as letting u=(x^2-1)^0.5 but that failed as well , and taylor expansion wouldn't help as it's an infinite series .

now I seem to be running out of methods .

I will appreciate any suggestion for an approach

Last edited by a moderator: May 7, 2017
2. Nov 15, 2014

### Staff: Mentor

The first thing I would try is combining the two fractions. I haven't worked the problem yet, so am not sure this will work. In any case, I would take that approach before doing anything like attempting to use L'Hopital's Rule.

BTW, is the "1" on the right side a typo? I don't get that the limit is 1.

Last edited by a moderator: May 7, 2017
3. Nov 15, 2014

### Staff: Mentor

Factor out an x2 from each of the radicals in the denominators. Then you will have 1's and 1/x2 within the radicals, and at large x, the 1/x2 will be small. Expand the resulting radicals using the binomial expansion. This will get rid of the radicals in each of the fractions. Then reduce to common denominator.

Chet

4. Nov 15, 2014

### throneoo

I have tried that , which resulted in failure . the fact that the limit is 1 seems unbelievable indeed

5. Nov 15, 2014

### throneoo

thanks , I seem to be able to evaluate the limit now .

the limit would then become x2 (1/sqrt(1-a) - 1/sqrt(1+a)) where a=1/x2 .

if I only expand sqrt(1+-a) up to order 1 , I would get 1/(1-1/2x4 , which would result in limit=1.

6. Nov 15, 2014

### Staff: Mentor

After a bit more work, I do get 1. In my erroneous work, I glossed over the fact that the numerator was approaching 0 while the x3 factor was getting large. In effect, I missed the [0 * ∞] indeterminate form.

Chet suggested one way. Another way will work as well. Combine the two fractions, then bring out factors of x from each radical. At that point, you can multiply by 1 in the form of the conjugate over itself. That's probably easier than doing the binomial expansion that Chet suggested.