1. ## Equation of tangent

Hello im stuck with a problem and the ONLY way i wanna solve it with a formel we suposed to use not derivate.
$Y=\frac{x-1}{x-2}$ on point (3,2)
the formel ima use is lim x->0 $\frac {f(a+h)-f(a)}{h}$and a=3
Progress:
$\frac{a+h-1}{a+h-1}-2/h$ it is suposed to mean that all that divide by h im trying my best with latex does not work well (notice i rewrite (3-1)/(3-2) as 2...
= $\frac{a+h-1-2a-2h+4}{a+h-2}/h$ (notice that h is suposed to divide all that i mean like (a/b)/(c/d) and h is the c and that in latex code is a at top and b the one down.
this is where i notice this dont work over and over... i am doing something wrong that i cant see

2. ## Re: Equation of tangent

I've been out of the game for a long time so you can check my algebra to see that it's correct.

If $f(x)=\frac{x-1}{x-2}$, then

$f(x+h)-f(x) = \frac{x+h-1}{x+h-2} - \frac{x-1}{x-2} = \frac{(x+h-1)(x-2) - (x-1)(x+h-2)}{(x+h-2)(x-2)} = \frac{x^2+xh-x-2x-2h+2-(x^2+xh-2x-x-h+2)}{(x+h-2)(x-2)} = \frac{-h}{(x+h-2)(x-2)}$

hence
$\frac{f(x+h)-f(x)}{h} = \frac{-1}{(x+h-2)(x-2)}$

now take the limit as h goes to zero to find the equation for the slope of a tangent curve where the function f(x) is continuous ie. wherever x is not equal to 2. so, plug in x=3 after evaluating the limit

Of course, this is by using definition alone, but it's way easier if you prove the general case of curves of f(x) = g(x)/h(x) (which would give you quotient rule, which is a special case of product rule. essentially re-prove product rule if you haven't in class can save some time) and then use substitution of g(x) = x-1 and h(x) = x-2.