cancel the h ... then let h approach 0.
Here is the question:
For the function
g(t)=2/t+3
Using the limit definition of the derivative find g'(t)
Ive gotten this far:
g'(t)=lim h->0 g(t+h) -g(t)/h
=lim h->0 2/t+h+3 - 2/t+3 all over h
=lim h->0 2t+6 -(2t+2h+6)/(t+3)(t+h+3) all over h
and I got that down to this:
-2h/(t+3)(t+h+3) all over h
This is where i am stuck. Help please!