Indeterminate limit involving (ax^2+b)^-0.5 and infinity

[throneoo]

Homework Statement

http://www5b.wolframalpha.com/Calculate/MSP/MSP19531i2efadg8b4hce2f00000hb7532fcd20a6h7?MSPStoreType=image/gif&s=47&w=253.&h=47.
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

 

[Mark44]

Homework Statement

http://www5b.wolframalpha.com/Calculate/MSP/MSP19531i2efadg8b4hce2f00000hb7532fcd20a6h7?MSPStoreType=image/gif&s=47&w=253.&h=47.
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

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.

 

[Chestermiller]

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

 

[throneoo]

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.

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

 

[throneoo]

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

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.

 

[Mark44]

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

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.