Determine the limit:
$$\lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}}$$
After taking the conjugate, I got:
$$\lim_{h\to 0} \frac{\big(\cos(x+h) - \cos(x)\big)\big((x+h)^{1/2} + x^{1/2})\big)} h$$
I took the conjugate of this, but I don't see how I can cancel out the $h$.
Any tips?
Hint:
You're supposed to spot derivatives here. Write your quotient as
$$\frac{\cos(x+h)-\cos x}{h}\cdot\frac{h}{(x+h)^{1/2}-x^{1/2}}$$
\begin{align} & \lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}} = \frac{\lim_{h\to0} \Big(\cos(x+h) - \cos x\Big) / h}{\lim_{h\to0} \Big( (x+h)^{1/2} - x^{1/2} \Big) / h} = \frac{\dfrac d {dx} \cos x }{\dfrac d {dx} x^{1/2}} = \cdots\cdots \end{align}
Simpler $$\frac{\cos(x+h)-\cos(x)}{(x+h)^{1/2} - x^{1/2}}\times\frac{(x+h)^{1/2} + x^{1/2}}{(x+h)^{1/2} + x^{1/2}}=\frac{\cos(x+h)-\cos(x)}{h}[(x+h)^{1/2} + x^{1/2}]$$
Hint :
$$\dfrac1{\sqrt{x+h}-\sqrt x}=\dfrac{x+h-x}{\sqrt{x+h}+\sqrt x}=?$$
and Use Prosthaphaeresis Formulas $$\cos(x+h)-\cos x=-2\sin\dfrac h2\sin\dfrac{2x+h}2$$
Finally, $\lim_{u\to0}\dfrac{\sin u}u=1$
