L'hopitals questions

Hey guys! I was hoping you could answer and explain a couple of these L'hopitals questions please:

Limit x -> infinity (x^2)/(sqrt(x^2+1))

Limit x-> 0 (sqrt(25-x^2)-5)/x

Limit x ->2+ ((8/(x^(2)-4))-(x/(x-2)))

Thank you very much in advance!

You don't actually need L'Hopital's for any of these.

1.) Note first that [x^2 / sqrt(x^2 + 1)]*[sqrt(x^2 + 1) / sqrt(x^2 + 1)] =

[x^2 / (x^2 + 1)] * sqrt(x^2 + 1). So

lim(x->infinity)[x^2 / sqrt(x^2 + 1)] = lim(x->infinity)[(x^2 / (x^2 + 1))*sqrt(x^2 + 1)] =

lim(x->infinity)[x^2/(x^2 + 1)] * lim(x->infinity)(sqrt(x^2 + 1)) =

lim(x->infinity)[1 / (1 + (1/x^2))] * lim(x->infinity)(sqrt(x^2 + 1) = 1*infinity = infinity.

If we were to use L'Hopital's Rule, then take the derivative of top and bottom to get

lim(x->infinity)[2x / ((1/2)*(x^2 + 1)^(-1/2) * 2x)] =

lim(x->infinity)(2*sqrt(x^2 + 1)) = infinity.

2.) Note first that [(sqrt(25 - x^2) - 5) / x]*[(sqrt(25 - x^2) + 5) / (sqrt(25 - x^2) + 5)] =

[(25 - x^2) - 25] / [x*(sqrt(25 - x^2) + 5)] = -x / (sqrt(25 - x^2) + 5).

So lim(x->0)[(sqrt(25 - x^2) - 5) / x] = lim(x->0)[ -x / (sqrt(25 - x^2) + 5)] = -0/10 = 0.

If we were to use L'Hopital's Rule, then take the derivative of top and bottom to get

lim(x->0)[(1/2)*(25 - x^2)^(-1/2) * -2x / 1] = lim(x->0)(-x / sqrt(25 - x^2)) = -0/5 = 0.

3.) Note first that (8/(x^2 - 4)) - (x/(x - 2)) = (8/(x^2 - 4)) - [x*(x+2) / ((x-2)(x+2))] =

(8 - x*(x+2)) / (x^2 - 4) = (8 - 2x - x^2) / (x^2 - 4). Now as x -> 2+ this goes to

the indeterminate form -0/0, so we can use L'Hopital's Rule:

lim(x->2+)[(8 - 2x - x^2)/(x^2 - 4)] = lim(x->2+)[(-2 - 2x)/(2x)] =

lim(x->2+)[-(x + 1) / x] = lim(x->2+)[-(1 + (1/x)] = -(1 + (1/2)) = -3/2.