but instead of f(x)=x^-2, let f(x) = x^(1/2 )
So, I did
[ (x+h)^1/2 - (x)^1/2 ]/ h
But im not sure how to simplify it. What would be my next step? Thanks alot
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}
$
$\displaystyle \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}
$
$\displaystyle \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$
finish it.
I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Reso...principles.pdf
If you have any questions we can answer them here
$\displaystyle \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}$
$\displaystyle = \lim_{h \to 0} \frac{2}{2} \times \frac{sin(2x+2h)-sin(2x)}{h} $
$\displaystyle = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}$
$\displaystyle = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H}$ where X=2x and H=2h
$\displaystyle = 2cos(X)$ ; which you proved before doing this.
$\displaystyle = 2cos(2x)$