Hello jzellt!

Did you ever hear about the Rule of L'Hospital?

a) lim (x->0) sinx/x^2

lim (x->0) sinx = 0 and lim (x->0) x^2 = 0

So [sin(x)]' = cos(x) and [x^2]' = 2x

Thus lim (x->0) sinx/x^2 = lim (x->0) cosx/(2x)

It is lim (x->0) cosx = 1 and lim(x->0)2x = 0

So

lim (x->0) cosx/(2x) =

b) lim (x->3) (x+3) / (x^2 - 9)

You want to find the limes of +3? You are sure about that? In that case it is

lim (x->3) (x+3) / (x^2 - 9) =

because (x+3) -> 6, if x->3

and x^2-9 -> 0, if x -> 0

Regards

Rapha