This has been bugging me all day, I feel like I am doing it right, but alas, my answer is always wrong

$\displaystyle

y = \frac{1}{\sqrt{x}}

$

Find the equation of the tangent line,f(x), at the point (4, 1/2)

$\displaystyle

\lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{4}}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{4}}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{\frac{2-\sqrt{x}}{2\sqrt{x}}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{\frac{2-\sqrt{x}(2+\sqrt{x})}{2\sqrt{x}(2+\sqrt{x})}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{\frac{4-x}{4\sqrt{x}+2x}}{x-4}

$

=

$\displaystyle

\lim_{x \to 4} \frac{4-x}{4\sqrt{x}+2x} * \frac{1}{x-4}

$

=

$\displaystyle

$$\displaystyle \lim_{x \to 4} \frac{-1}{4\sqrt{x}+2x}

$

=

$\displaystyle

$$\displaystyle \lim_{x \to 4} \frac{-1}{4\sqrt{4}+2(4)}

$

=

$\displaystyle

\frac{-1}{16}

$

is that right? (as far as the slope is concerned)

and the equation would be $\displaystyle y-.5 = \frac{-1}{16}(x-4) $

$\displaystyle

y= \frac{-1}{16}x+.75

$