Evaluate the following limits
(i) `lim_(x to (pi)/(2)) tan^(2) x [sqrt(2 sin^(2) x + 3 sin x + 4) - sqrt(sin^(2) x + 6 sin x + 2)]`
(ii) `lim_(theta to 0) (sqrt(1 + sin 3 theta) -1)/(ln(1 + tan 2theta))`
(iii) `lim_(x to 0) (sqrt(1 + x) - ""^(3)sqrt(1 + x))/(x)`
(iv) `lim_(phi to 0) (8)/(phi^(8)) (1 - cos (phi^(2))/(2) - cos (phi^(2))/(4) + cos (phi^(2))/(2). cos (phi^(2))/(4))`

Explanation

We have,
`underset(x to (pi)/(2))(lim) tan^(2) xx [sqrt(sin^(2) x + 3 sin x + 4) - sqrt(sin^(2) x + 6 sin x + 2)]`
`underset(x to (pi)/(2))(lim) (sin^(2)x)/(cos^(2)x) [(2sin^(2) x + 3sin x + 4 - sin^(2) x -6 sin x - 2)/(sqrt(2 sin^(2) x + 3 sin x + 4 ) + sqrt(sin^(2)x + 6 sin x + 2))]`
`underset(x to (pi)/(2))(lim) (sin^(2) x (sin^(2) x - 3 sin x + 2))/((1 - sin x)(1 + sin x)[sqrt(2 sin^(2) x + 3 sin x + 4) + sqrt(sin^(2) x 6 sin x + 2)])`
`= underset(x to (pi)/(2))(sin^(2) x (sin x - 1)(sin x - 2))/((1 - sin x)(1 + sin x)[sqrt(2 sin^(2) x + 3 sin x + 4 )+sqrt(sin^(2) x + 6 sin x + 2)])`
`= underset(x to (pi)/(2))(sin^(2) x (sin x - 2))/((1 + sin x)[sqrt(2 sin^(2) x + 3 sin x + 4) + sqrt(sin^(2) x + 6 sin x + 2)])`
`= (-1(1-2))/(2(3 + 3)) = (1)/(12)`
We have,
`underset(theta to 0)(lim) (sqrt(1 + sin theta)-1)/(In (1 + tan 2 theta))`
Multiply numerator and denominator by `sqrt(1 + sin 3 theta) + 1` we get
`underset(theta to 0)(lim) (sqrt(1 + sin theta) - 1)/(In (1 + tan 2 theta)) = underset(theta to 0)(lim) (1 + sin 3 theta - 1)/([sqrt(1 + sin 3 theta) + 1][In (1 + tan 2 theta)])`
`underset(theta to 0)(lim) ((sin 3 theta0)/(3 theta) 30)/([sqrt(1 + sin 3 theta + 1)(In(1 + tan 2 theta ))/(tan 2 theta)(tan 2 theta)/(2 theta)]) 2 theta`
` = (3)/(sqrt(2) xx 1 xx 1 xx 2) = (3)/(2sqrt(2))`
We have,
`underset(x to 0)(lim) (sqrt(1 + x) - .^(3)sqrt(1 + x))/(x)`
Let us set `(1 + x)^(1//6) = t` so that `t^(6) = 1 + x` and `x to 0 implies t to 1` with this substitution the given limit reduce to
`underset(x to 0)(lim) (sqrt(1 + x) - .^(3)sqrt(1 + x))/(1 + x = 1) = underset(t to 1)(lim) (t^(3) - t^(2))/(t^(6) - 1) = underset(t to 1)(lim) (t^(2) (t - 1))/((t^(6) - 1)/(t - 1) xx (t - 1))`
`= (1)/(6) ( :' underset(x to 1)(lim) (x^(n) - 1)/(x - 1) = n)`
we have,
(iv) We have,
`underset(phi to 0)(lim) (8)/(phi^(8)) (1-cos (phi^(2))/(2) - cos (phi^(2))/(4) + cos (phi^(2))/(2) cos (phi^(2))/(4))`
`underset(phi to 0)(lim) (8)/(phi^(8)) (1 - cos (phi^(2))/(2))(1 - cos (phi^(2))/(4))`
`underset(phi to 0)(lim) (8)/(phi^(8)) 2 sin^(2) (phi^(2))/(4) 2 sin^(2) (phi^(2))/(8)`
`underset(phi to 0)(lim) (32)/(phi^(8)) ((sin (phi^(2))/(4))/((phi^(2))/(4)))^(2) ((sin(phi^(2))/(8))/((phi^(2))/(8)))^(2) (phi^(4))/(16). (phi^(4))/(64)`
`= 32 xx 1^(2) xx 1^(2) xx (1)/(16) xx (1)/(64) = (1)/(32)`