Calculus help?

Question

First person to answer both correctly gets the 10 points.

lim t->-2 (1/2+1/t)/(2+t)

and

lim x->0 (tan2x)/(5x)

Answer 1

lim t->-2 (1/2+1/t)/(2+t)
Multiply top and bottom by 2t:
= lim t->-2 (t + 2) / [(2t) (2+t)]
Cancel common factors:
= lim t->-2 1 / 2t
= -1/4.

lim x->0 (tan 2x) / (5x)
Use L'Hopital's Rule here: (you can also get it using lim x->0 sin x / x = 1 if you have that but not L'Hopital's)
= lim x->0 (2 sec^2 2x) / 5
= (2 . 1^2) / 5
= 2/5.

Answer 2

1)let t=x-2
lim x->0 (1/2+1/x-2)/x
since both nr dr ->0
using l'hospital rule
lim x->0 -1/(x-2)^2 /1=-1/4

2)lim x->0 (tan 2x)/(5x)
similar case
lim x->0 sec^2 2x * 2/5=(2/5)sec^2 2x
lim x->0 1*(2/5)=2/5
so ans[=2/5