Finding Tangent Line Using Limit Definition I'm supposed to get the equation of the tangent line to the graph of $f(x)= \frac{8}{x}$ at the point $(2,4)$.
I started with
$$\frac{\frac{8}{x+h} - \frac{8}{x}}{h},$$
then I cross multiplied: 
$$\frac{8x - 8(x+h)}{
x(x+h)}$$
Where do I go from here?
 A: Then you divide by $h$.
Using, as $h \to 0$,
$$
\frac{f(x+h)-f(x)}{h} \to f'(x)
$$ with $f(x)=\dfrac8{x}$, $x \neq0$, gives, as $h \to 0$,
$$
\frac{\dfrac8{x+h}-\dfrac8{x}}{h}=\frac{\dfrac{8x-8x-8h}{x(x+h)}}{h}=\frac{\dfrac{-8h}{x(x+h)}}{h}=-\dfrac{8}{x(x+h)} \to -\dfrac{8}{x^2},
$$ and
$$
f'(x)=-\dfrac{8}{x^2}.
$$ Then you may use the equation of the tangent line

$$
y-f(x_0)=f'(x_0)(x-x_0)
$$ 

to conclude.
A: Actually you forgot the $h$ term in the denominator. We have $\frac{8x-8(x+h)}{x(x+h)h}=\frac{-8h}{x(x+h)h}$ and this simplifies to $\frac{-8}{x(x+h)}$. We need to take the limit of this term as $h \rightarrow 0$. This is simply $\frac{-8}{x^2}$, which is exactly the derivative of the curve.
Evaluate the expression $\frac{-8}{x^2}$ at $x=2$ for the exact numerical value of the slope of the tangent at the given point. Then apply the fact that a straight line is uniquely determined by its slope and a point it passes through using the equation $(y-y_0)=m \times (x-x_0)$, where $m$ is the slope, which we computed earlier, and $(x_0, y_0)$ is the point you already mentioned in the question. Does that make sense?
