Calculus--Limits Help!!?

Question

Okay I really need some help with my Calculus homework I don't get any of these problems:

1) lim x->0 tan^2x / x

2) lim x-> pi x sec x

3) lim x-> pi/4 1-tanx / sinx-cosx

4) lim x-> 0 cosx-1 / 2x^2

5) lim x->0 sinx / cube rt. of x

I would really appreciate your help on any of these problems, please show steps so I understand what you are doing.

Answer 1

lim (x->0) tan^2 x / x: this is in 0/0 form, so you can use L'Hopital's rule:
= lim (x->0) 2 tan x sec^2 x / 1
= 2 (0) (1)^2 / 1
= 0.

lim (x->π) x sec x = π sec π
= π / cos π
= -π.

lim (x->π/4) (1 - tan x) / (sin x - cos x)
= lim (x->π/4) (cos x - sin x) / [cos x (sin x - cos x)]
= lim (x->π/4) (-1) / cos x
= -1/√2.

lim (x->0) (cos x - 1) / 2x^2; 0/0 form, use L'Hopital's Rule:
= lim (x->0) (-sin x) / 4x; 0/0 form again, L'Hopital again:
= lim (x->0) (-cos x) / 4
= -1/4.

Note you can also get this from the expansion of
cos x = 1 - x^2 / 2! + x^4 / 4! - x^6 / 6! + ...
so (cos x - 1) / 2x^2 = (-x^2 / 2 + x^4 / 24 - x^6 / 720 + ...) / 2x^2
= (-1/4 + x^2 / 48 - x^4 / 1440 + ...)
-> -1/4 as x -> 0.

lim (x->0) sin x / x^(1/3); 0/0 form, use L'Hopital's Rule:
= lim (x->0) cos x / (1/3 x^(-2/3))
= lim (x->0) 3 x^(2/3) cos x
= 3 (0) (1)
= 0.

Answer 2

Damn Im studying this in a few months