All i ask is how to write it, I can figure the answer myself. So far I think its like this:

lim h->0 2(x+3+h)-2x+3/h; however I'm not confident in that answer because all the x variables will be gone and with h=0 the answer be nothing. Please correct me where I am interpreting this wrong.

If f(x) =2x+ 3 then f(x+h) = 2(x+h) + 3 not 2(x+3+h)

you obtain

lim (2h/h) = lim2 = 2 be thankful for how easy it simplified

For f(x)=2x^2+3x-2

f(x+h) = 2(x+h)^2 + 3(x+h) -2

and now you wish it were as simple as the first

lim { [ 2(x+h)^2 + 3(x+h) -2] -[ 2x^2+3x-2]}/h