TITLE: Finding $\lim\limits_{h \to 0}\frac{(\sqrt{9+h} -3)}{h}$
QUESTION [6 upvotes]: I have about 10 of these problems I could post, can't figure any of them out...not sure what to do.
$$\lim_{h \to 0}\dfrac{(\sqrt{9+h} -3)}{h}$$
I know that I need to rationalize it so I multiply by the numerator which gives me 
$9 + h - 9$ (not sure if positive or negative since it is $-3$ squared it should be $9$ but it could also be minus square of $3$ which would be -9
/h sqr(9+h) -3 
From that I can divide by h and get 1/3-3
which I know is wrong. What am I doing wrong?

REPLY [2 votes]: Posting as I havent seen anyone suggest Taylor expansion yet.
$$\lim_{h \to 0}\dfrac{\sqrt{9+h} -3}{h} = \lim_{h \to 0}\dfrac{3\sqrt{1+h/9} -3}{h}$$
Taylor expanding the numerator becomes  $3(1+ \frac{1}{2}\cdot\frac{h}{9}) -3 = \frac{h}{6} + O(h^2)$
$$\lim_{h\to 0}
\dfrac{h/6+O(h^2)}{h} = \frac{1}{6}$$