Compute $f'(x)$ using the limit definition
$$f(x)=x-\sqrt { x } $$
Steps I took:
$$f'(x)=\lim _{ h\rightarrow 0 }{ \frac { x+h-\sqrt { x+h } -(x-\sqrt { x } ) }{ h } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \frac { x+h-\sqrt { x+h } -x+\sqrt { x } }{ h } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \frac { h-\sqrt { x+h } +\sqrt { x } }{ h } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad \frac { h }{ h } -\frac { \sqrt { x+h } +\sqrt { x } }{ h } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad 1-\frac { \sqrt { x+h } +\sqrt { x } }{ h } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad 1-(\frac { \sqrt { x+h } +\sqrt { x } }{ h } \cdot (\frac { \sqrt { x+h } -\sqrt { x } }{ \sqrt { x+h } -\sqrt { x } } )) } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad 1-\frac { x+h-x }{ h(\sqrt { x+h } -\sqrt { x } ) } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad 1-\frac { h }{ h(\sqrt { x+h } -\sqrt { x } ) } } \quad $$
$$f'(x)=\lim _{ h\rightarrow 0 }{ \quad 1-\frac { 1 }{ \sqrt { x+h } -\sqrt { x } } } \quad $$
So I got to this point and realize that as $h$ approaches zero and gets cancelled out the denominator would become zero. My answer looks similar to what the correct answer should look like except for the denominator. Where did I go wrong? I would like a hint. No direct answer. That won't help me.
