Chapter 4 #3

Use the definition to find the derivative of $f(x)=\sqrt {x}$, for $x>0$.  Is $f$ differentiable at zero?  Explain.

$\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t}$

     $=\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t} * \dfrac {\sqrt {x+t}+\sqrt {x}}{\sqrt {x+t}+\sqrt {x}}$

     $=\lim_{t\rightarrow 0} \dfrac {x+t-x}{t[\sqrt {x+t}+\sqrt {x}]}$

     $=\lim_{t\rightarrow 0} \dfrac {1}{\sqrt {x+t}+\sqrt {x}}$

     $=\dfrac {1}{2\sqrt {x}}$

$f(x)=\sqrt {x}$ is not differentiable at $x=0$ because the limit of $T(x)=\dfrac {f(x)-f(x_0)}{x-x_0}$ as $x$ approaches zero does not exist.

$\lim_{x\rightarrow 0} \dfrac {\sqrt {x}-\sqrt {0}}{x-0}$

     $=\lim_{x\rightarrow 0} \dfrac {1}{\sqrt {x}}$

     $=\dfrac {1}{0}$ and is undefined.