Use analytical methods to evaluate the following limits.

Use analytical methods to evaluate the following limits.

1.) Limit as x approaches infinity of (√(x - 2) - √(x - 4))

2.) Limit as x approaches infinity of x^3((1/x) - sin(1/x))

3.) Limit as x approaches infinity of (√(x^2 - 1) - (x^3 - 1)^1/3)

4.) Limit as x approaches infinity of ((1/(x - 1)) - (1/(√(x-1))))

The first and third one need a special "trick" to evaluate. The second and last one can be done by combining fractions and doing some cancellations.

(1) Multiplying √(x - 2) - √(x - 4) by 1 = [√(x - 2) + √(x - 4)]/[√(x - 2) + √(x - 4)] gives:
lim (x-->infinity) [√(x - 2) - √(x - 4)]
= lim (x-->infinity) [√(x - 2) + √(x - 4)[√(x - 2) - √(x - 4)]]/[√(x - 2) + √(x - 4)]
= lim (x-->infinity) [(x - 2) - (x - 4)]/[√(x - 2) + √(x - 4)], by difference of squares
= lim (x-->infinity) 2/[√(x - 2) + √(x - 4)].

The denominator goes to infinity as x --> infinity, so this goes to 0.

(2) can be done combining fractions and using L'Hopital's Rule. It shouldn't be that bad.

(3) requires that we use the rationalization track we did in (1) TWICE: once to rationalize √(x^2 - 1) and again to rationalize (x^3 - 1)^(1/3).

Multiplying √(x^2 - 1) - (x^3 - 1)^(1/3) by [√(x^2 - 1) + (x^3 - 1)^(1/3)]/[√(x^2 - 1) - (x^3 - 1)^(1/3)] will rationalize √(x^2 - 1) using difference of squares.
√(x^2 - 1) - (x^3 - 1)^(1/3)
= [√(x^2 - 1) - (x^3 - 1)^(1/3)][√(x^2 - 1) + (x^3 - 1)^(1/3)]/[√(x^2 - 1) + (x^3 - 1)^(1/3)]
= [(x^2 - 1) - (x^3 - 1)^(2/3)]/[√(x^2 - 1) + (x^3 - 1)^(1/3)].

Then, multiplying the numerator and denominator by:
(x^2 - 1) + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3),

will allow us to write:
(x^2 - 1)^3 - (x^3 - 1)^2 = [(x^2 - 1) - (x^3 - 1)^(2/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)],

using the difference of cubes formula:
a^3 - b^3 = (a - b)(a^2 + ab + b^2).

So, putting this altogether gives: