find the limit when x approaches 0

Question

johnnydicamillo · Answer

$$\lim_{x \rightarrow 0}\frac{ \sqrt{9 + h} - 3 }{ h }$$

johnnydicamillo · Answer

I know I need to rationalize the numerator but I think I am doing some mathematical error. Can someone show me how to rationalize the numerator?

geerky42 · Answer

You mean $$\large \lim_{h\to0}\dfrac{\sqrt{9+h}-3}{h}$$ right?

You should rationalize numerator, then simplify it. Works for me.

What went wrong?

geerky42 · Answer

To rationalize numeration, you just multiply numerator and denominator by conjugate of numerator: $\sqrt{9+h}+3$

geerky42 · Answer

numerator *

johnnydicamillo · Answer

and I get $$\frac{ 9 + h - 9 }{ h \sqrt{9 + h} + 3 }$$

johnnydicamillo · Answer

but it the numerator would still end up being 0

geerky42 · Answer

Try to simplify.

freckles · Answer

and the sqrt(9+h)+3 should have ( ) around it (on bottom)

johnnydicamillo · Answer

okay, um... give me a sec

johnnydicamillo · Answer

i am stuck

geerky42 · Answer

You have $\dfrac{9+h-9}{h(\sqrt{9+h}+3)}$, right?
$$\dfrac{9+h-9}{h(\sqrt{9+h}+3)}\~\\dfrac{h}{h(\sqrt{9+h}+3)}\~\\dfrac{\cancel h}{\cancel h(\sqrt{9+h}+3)}\~\\dfrac{1}{\sqrt{9+h}+3}$$

johnnydicamillo · Answer

got, it.. For some reason the h just disappeared in my head. Thanks

geerky42 · Answer

No problem