How to use the alternate definition to find the derivative of ##f(x)=sqrt(x+3)## at x=1?

Here are a couple ways you can do the limit calculation for the derivative. Both methods involve "rationalizing the numerator" (not the denominator) as a trick to help you calculate the limits.

##f'(1)=lim_{h->0}\frac{f(1+h)-f(1)}{h}##

##=lim_{h->0}\frac{\sqrt{4+h}-2}{h}\cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}##

##=lim_{h->0}\frac{4+h-4}{h(\sqrt{4+h}+2)}=lim_{h->0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{4}##

OR

##f'(1)=\lim_{x->1}\frac{f(x)-f(1)}{x-1}=\lim_{x->1}\frac{\sqrt{x+3}-2}{x-1}##

##=lim_{x->1}\frac{x+3-4}{(x-1)(\sqrt{x+3}+2)}##

##=lim_{x->1}\frac{1}{sqrt{x+3}+2}=\frac{1}{4}##