SIMPLE calculus limit question PLEASE answer I vote fast :)

i don't understand how this problem and i really should.

limT-->0 T /(square root(4+T)- square root(4-T))

how do you solve this damn thing!

and also shouldn't

lim T-->0 (T^2)(3T) / ((T+2)^2 - (T-2)^2)

shouldn't this limit not exist because the denominator is equal to zero...my answer book gives me 3/8 but that would have to mean the denominator is 4 + 4 instead of 4 - 4

please tell me where I am going wrong
thank you!

Any time you get 0/0 by substitution, you have to rewrite the function, either by factoring, multiplying by the conjugate, etc...
A) limT-->0 T /(square root(4+T)- square root(4-T))

Lim T/[sqr(4+T)-sqr(4-T)] = 0/0

= T*[sqr(4+T)+sqr(4-T)]......... multiply by conjugate of the denominator.
-----------------------------
[sqr(4+T)-sqr(4-T)] [sqr(4+T)+sqr(4-T)]

=T[sqr(4+T)+sqr(4-T)]
-------------------------- Notice that the denominator is 2T, so cancel T
[(4+T)-(4-T)]

= [sqr(4+T)+sqr(4-T)]
--------------------------
2

Now plug in 0

=>4/2=2

B) lim T-->0 (T^2)(3T) / ((T+2)^2 - (T-2)^2) =0/0

= 3T^3
-------------------
(T+2)^2-(T-2)^2

=3T^3
-------------------
(T^2+4T+4)-(T^2-4T+4)

=3T^3
-------
8T

=(3/8)T^2

->0

Maybe the t^2 wasn't part of the problem? In that case, it simplifies to 3/8

Hoping this helps!

lim t/[√(4+t) -√(4-t) = lim t[√(4+t) +√(4-t)]/(4+t-4+t)
t→0 t→0
=lim t[√(4+t) +√(4-t)]/2t
t→0
=lim [√(4+t) +√(4-t)]/2
t→0
= (√4 +√4)/2
=2

the only way that the answer is 3/8 is

lim 3t/[(t+2)²-(t-2)²] =lim 3t/(t²+4t+4-t²+4t-4)
t→0 t→0
=lim 3t/8t
t→0
= 3/8