Question

Evalute `lim_(x->oo) [sqrt(x^2+x+2) - sqrt(x^2-3x-5)]`?

Answer 1

`lim_(x->oo) sqrt(x^2+x+2) - sqrt(x^2-3x-5)`

`= lim_(x->oo) x ( sqrt(1+1/x+2/x^2) - sqrt(1-3/x-5/x^2))`

This next bit is an algebraic shortcut as the `1/x^2` terms seem to disappear, but those terms are going to be order of magnitudes smaller. We could do a full expansion but that would look quite messy. So, from there:

`= lim_(x->oo) x ( sqrt(1+1/x) - sqrt(1-3/x))`

By the binomial series expansion, `(1+ z)^alpha = 1 + alpha z + O(z^2)`:

`= lim_(x->oo) x ( (1+1/(2x) + O(1/x^2)) - (1-3/(2x) + O(1/x^2) )`

`= lim_(x->oo) x * (1/(2x) + 3/(2x)) = 2`

Answer 2

2

Explanation:

`lim_(x->oo) [sqrt(x^2+x+2) - sqrt(x^2-3x-5)]`
`lim_(x->oo) [{(sqrt(x^2+x+2) - sqrt(x^2-3x-5))(sqrt(x^2+x+2) + sqrt(x^2-3x-5))}/(sqrt(x^2+x+2) + sqrt(x^2-3x-5))]`
`lim_(x->oo) [(sqrt(x^2+x+2))^2 - (sqrt(x^2-3x-5))^2]/(sqrt(x^2+x+2) + sqrt(x^2-3x-5))`
`lim_(x->oo) [{(x^2+x+2-x^2+3x+5)}/(sqrt(x^2+x+2) + sqrt(x^2-3x-5))]`
`lim_(x->oo) [(4x+7)/(sqrt(x^2+x+2) + sqrt(x^2-3x-5))]`
`lim_(x->oo) [{(4x+7)/x}/{(sqrt(x^2+x+2) + sqrt(x^2-3x-5))/x}]`
`lim_(x->oo) [{4+(7/x)}/{(sqrt(1+1/x+2/x^2))+(sqrt(1-3/x-5/x^2))}]`

As we know:
`(lim_(x->oo)1/x = 0) and (lim_(x->oo)1/x^2 = 0)`

Therefore:
`= [(4+0)/(sqrt(1+0+0)+sqrt(1-0-0))]`
`= [4/2]`
`= 2`