What is the lim of h approaching 0 of (3^2-(3+h)^2)/(h) and how do you solve it?

Question

zepdrix · Answer

Was the problem presented like this?
Because your numerator looks backwards.

zepdrix · Answer

$$\Large\rm \lim_{h\to0}\frac{(3+h)^2-3^2}{h}$$This would be the limit definition of the derivative of the function $\Large\rm f(x)=x^2$ evaluated at x=3.

anonymous · Answer

$$\frac{ 3^{2}-(3+h)^{2} }{ h }$$

anonymous · Answer

$$\lim_{h \rightarrow 0}$$

zepdrix · Answer

Alright fine :O If you say so...

So expand out your (3+h)^2.
Remember how to do that?$$\Large\rm (3+h)^2=(3+h)(3+h)=?$$

anonymous · Answer

so that would =9+6h+h^2

zepdrix · Answer

$$\Large\rm \lim_{h\to0}\frac{3^2-(9+6h+h^2)}{h}$$Ok good.
Understand why I put it in brackets?
It's important that you distribute the subtraction to each of the three terms.

anonymous · Answer

yes, so i have changed the signs to $$\lim_{h \rightarrow 0}\frac{ 3^{2}+9-6h-h ^{2} }{ h }$$
that do i do with the $$3^{2}$$

zepdrix · Answer

Square the 3 silly 0_o

zepdrix · Answer

Woops you didn't distribute the minus to the 9.

anonymous · Answer

Oh woops, so the 9's would cross out leaving $$\frac{ -6h-h ^{2} }{ h }$$ and i understand how to find the rest of the problem. Thank you very much!!

zepdrix · Answer

yay team!