Use analytical methods to evaluate the following limits.
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Use analytical methods to evaluate the following limits.

[From: ] [author: ] [Date: 12-03-25] [Hit: ]
Therefore, this limit is 0.(4) can be done by combining fractions (the LCD is x - 1).I hope this helps!-1) lim(x→∞) (√(x - 2) - √(x - 4))= lim(x→∞) (√(x - 2) - √(x - 4)) * (√(x - 2) + √(x - 4)) /(√(x - 2) + √(x - 4)), via conjugates= lim(x→∞) ((x - 2) - (x - 4)) /(√(x - 2) + √(x - 4))= lim(x→∞) 2/(√(x - 2) + √(x - 4))= 0.......
lim (x-->infinity) [√(x^2 - 1) - (x^3 - 1)^(1/3)]
= lim (x-->infinity) [(x^2 - 1) - (x^3 - 1)^(2/3)]/[√(x^2 - 1) + (x^3 - 1)^(1/3)]
= lim (x-->infinity) [(x^2 - 1)^3 - (x^3 - 1)^2]/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}
= lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}.

Now, at this point, we can use the fact that for x ≠ 0:
(a) √(x^2 - 1) - (x^3 - 1)^(1/3) = √[x^2(1 - 1/x^2)] - [x^3(1 - 1/x^3)]^(1/3)
= √(x^2)√(1 - 1/x^2) - (x^3)^(1/3)(1 - 1/x^3)^(1/3)
= x√(1 - 1/x^2) - x(1 - 1/x^3)^(1/3).
(b) (x^2 - 1)^2 - (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)
= [x^2(1 - 1/x^2)]^2 - [x^2(1 - 1/x^2)][x^3(1 - 1/x^3)]^(2/3) + [x^3(1 - 1/x^3)]^(4/3)
= x^4(1 - 1/x^2)^2 - x^4(1 - 1/x^3) + x^4(1 - 1/x^3)^(4/3).

Putting this altogether gives:
lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[√(x^2 - 1) + (x^3 - 1)^(1/3)][(x^2 - 1)^2 + (x^2 - 1)(x^3 - 1)^(2/3) + (x^3 - 1)^(4/3)]}
= lim (x-->infinity) (-3x^4 + 2x^3 + 3x^2 - 2)/{[x√(1 - 1/x^2) + x(1 - 1/x^3)^(1/3)][x^4(1 - 1/x^2)^2 - x^4(1 - 1/x^3) + x^4(1 - 1/x^3)^(4/3)]}
= lim (x-->infinity) (-3 + 2/x + 3/x^2 - 2/x^4)/{[x√(1 - 1/x^2) + x(1 - 1/x^3)^(1/3)][(1 - 1/x^2)^2 - (1 - 1/x^3) + (1 - 1/x^3)^(4/3)]}, by dividing the numerator and denominator by x^4.

The numerator goes to -3, while the denominator goes to infinity. Therefore, this limit is 0.

(4) can be done by combining fractions (the LCD is x - 1).

I hope this helps!

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1) lim(x→∞) (√(x - 2) - √(x - 4))
= lim(x→∞) (√(x - 2) - √(x - 4)) * (√(x - 2) + √(x - 4)) / (√(x - 2) + √(x - 4)), via conjugates
= lim(x→∞) ((x - 2) - (x - 4)) / (√(x - 2) + √(x - 4))
= lim(x→∞) 2/(√(x - 2) + √(x - 4))
= 0.

2) lim(x→∞) x^3 (1/x - sin(1/x))
= lim(t→0+) (1/t)^3 (t - sin t), letting t = 1/x
= lim(t→0+) (t - sin t)/t^3
= lim(t→0+) (1 - cos t)/(3t^2), by L'Hopital's Rule
= lim(t→0+) sin t/(6t), by L'Hopital's Rule again
= lim(t→0+) cos t/6, by L'Hopital's Rule again!
= 1/6.

3) lim(x→∞) (√(x^2 - 1) - (x^3 - 1)^(1/3))
= lim(x→∞) [√(x^2 (1 - 1/x^2)) - (x^3 (1 - 1/x^3))^(1/3)]
= lim(x→∞) [x √(1 - 1/x^2) - x(1 - 1/x^3)^(1/3)]
= lim(x→∞) [√(1 - 1/x^2) - (1 - 1/x^3)^(1/3)] / (1/x)
= lim(t→0+) [√(1 - t^2) - (1 - t^3)^(1/3)] / t, letting t = 1/x
= lim(t→0+) [-t(1 - t^2)^(-1/2) + t^2(1 - t^3)^(-2/3)] / 1, by L'Hopital's Rule
= 0.

4) lim(x→∞) [1/(x - 1) - 1/√(x - 1)] = 0 - 0 = 0.

I hope this helps!

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