Please Help !!!! evaluate the limit, if it exists.?

Question

lim [ (1 / t ) - ( 1 / ( t^2 + t ) ) ]
t-->0

____


lim (sqrt ( 28 - x ) - 5) / (sqrt (19 - x ) - 4)
x-->3

______

Answer 1

lim (t-> 0) (1/t - 1/(t^2+t))
t^2 + t = t(t+1), so this is the common denominator:
= lim (t-> 0) [((t+1) - 1) / (t (t+1))]
= lim (t-> 0) t / [t (t+1)]
= lim (t-> 0) 1 / (t+1)
= 1.

lim (x->3) (√(28-x) - 5) / (√(19-x) - 4)
Since this is in 0/0 form, use L'Hopital's Rule:
= lim (x->3) [(1/2) (28-x)^(-1/2) (-1)] / [(1/2) (19-x)^(-1/2) (-1)]
= lim (x->3) [(28 - x) / (19 - x)]^(-1/2)
= (25 / 16)^(-1/2)
= √(16/25)
= 4/5.

Answer 2

1. lim [ (1 / t ) - ( 1 / ( t^2 + t ) ) ]
t-->0
(( t^2 + t ) / t( t^2 + t ) ) - ( t / t( t^2 + t ) )
(( t^2 + t )-t)/( t( t^2 + t ))
t^2/( t( t^2 + t ))
t/( t^2 + t )
t/t(t+1)
1/(t+1)
1/(0+1)
1
2. lim (sqrt ( 28 - x ) - 5) / (sqrt (19 - x ) - 4)
x-->3
4/5

Answer 3

lt (1/t)(1-(1/1+t))
t-->0

=>lt (t+1-1)/(t(t+1))
t-->0

=> lt 1/(t+1)
t-->0

=1

Sorry, I dont know how to do the second.