F(x) = (x+1)/(x-1) a=3
Compute derivative at x = a using the limit definition and find an equation of the tangent line.
I get up to:
[(3+h)+1/(3+h)-1] - 2
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h
but I don't know where to go from there, help!
I'll start at the start. First of all, you should make life as easy for yourself as possible.
So write $\displaystyle \frac{x + 1}{x - 1}$ as $\displaystyle \frac{(x - 1) + 2}{x - 1} = 1 + \frac{2}{x - 1}$.
I'll do it for general a. If you prefer, you can replace all the a's with 3 .......
Note that if f(x) = 1, then $\displaystyle \frac{f(a + h) - f(a)}{h} = 0$ since f(x) = 1 for all values of x including x = a + h and x = a.
So you only have to worry about getting $\displaystyle \lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h}$ for $\displaystyle f(x) = \frac{2}{x - 1}$:
$\displaystyle f(a + h) - f(a) = \frac{2}{a + h - 1} - \frac{2}{a - 1}$
Put over a common denominator:
$\displaystyle = \frac{2(a - 1)}{(a + h - 1)(a - 1)} - \frac{2(a + h - 1)}{(a - 1)(a + h - 1)} = \frac{2(a - 1) - 2 (a + h - 1)}{(a + h - 1)(a - 1)}$
Simplify:
$\displaystyle = \frac{-2h}{(a + h - 1)(a - 1)}$.
So $\displaystyle \frac{f(a + h) - f(a)}{h} = \frac{-2}{(a + h - 1)(a - 1)}$.
Now take the limit
$\displaystyle \lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h} = \lim_{h \rightarrow} \frac{-2}{(a + h - 1)(a - 1)} = \frac{-2}{(a - 1)^2}$.
When a = 3 the value this limit is equal to $\displaystyle -\frac{1}{2}$.
This means that the gradient of the tangent line is $\displaystyle -\frac{1}{2}$.