Some limit problems that I dont understand?

Question

How do you solve this using the L'Hopital's Rule?
1. the limit of cotx/ln x as x approaches zero from the right side (0+)
2. the limit of cos((1/t))^2 as t approaches infinity


and as addition, what is the limit of
x + sin(2x) as x approaches infinity?

thanks all

Answer 1

1. the limit of cotx/ln x as x approaches zero from the right side (0+)
lim x → 0+ (cotx/lnx)
= lim x → 0+ (-cosec²x/(1/x)) (by l;Hôpital's Rule)
= lim x → 0+ (-x/sin²x)
= lim x → 0+ (-x/sinx . 1/sinx)
= - ∞ (-x/sinx→-1 while 1/sinx→1/0)

2. the limit of cos((1/t))^2 as t approaches infinity
lim t →∞ (cos((1/t))^2) = 1
since as t →∞, (1/t))^2) → 0 and cos 0 = 1

lim x →∞ (x + sin(2x) ) = ∞ as even though sin2x oscillates between -1 and 1, x →∞

Answer 2

) just differentiate cotx and ln x until you dont have 0/0 or inf/inf
if you have a/0 you cant use l'hopital.

2) idem dito as 1. replace 1/t t-> inf with 1/t =x and x ->0

x + sin(2x) x->inf
this limoit doesnt exist since sin(2x) x -> inf oscilates between +1 and -1

Answer 3

dont have my abstract books with me and its late but I remember Lhopitals rule states that f(x) / g(x) approches the same lim as f'(x) / g'(x)

Answer 4

I'll get back to you on that one...