# How do you use the limit definition to find the slope of the tangent line to the graph f(x)= x(sqrt(x)-1)  at x=4?

Nov 8, 2016

graph{(x(sqrt(x)-1)-y)(2x-y-4)=0 [-1.99, 13.814, -0.53, 7.37]}

$2 x - y - 4 = 0$

#### Explanation:

The tangent is given by

$y - f \left({x}_{0}\right) = {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} \cdot \left(x - {x}_{0}\right)$

Then

x_0=4\ \ \ \ ; \ \ \ f(x_0)=4

${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} = {\lim}_{h \to 0} \frac{\left(4 + h\right) \left(\sqrt{4 + h} - 1\right) - 4}{h} =$
$= {\lim}_{h \to 0} \left[\sqrt{4 + h} - 1 + 4 \frac{\sqrt{4 + h} - 2}{h}\right] =$
$= 2 - 1 + 4 {\lim}_{h \to 0} \frac{4 + h - 4}{h \cdot \left(\sqrt{4 + h} + 2\right)} = 2 - 1 + 4 \cdot \frac{1}{4} = 2$

So the tangent is

$y - 4 = 2 \left(x - 4\right)$