Another equation of a tangent

[teneleven]

Homework Statement

Find an equation of the tangent line to the curve at the given point.

$$y = \frac{x - 1}{x - 2}$$ , (3, 2)

Homework Equations

$$m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}$$

$$y = mx + b$$

The Attempt at a Solution

$$\lim_{h\to 0} \frac{\frac{(x + h) - 1}{(x + h) - 2} - \frac{x - 1}{x - 2}}{h}$$

$$\lim_{h\to 0} \frac{\frac{2 + h}{1 + h} - \frac{2}{1}}{h}$$

$$\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2}{h}$$

$$\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2 + 2h}{h + h^2}$$

$$\lim_{h\to 0} \frac{h + 2h}{h + h^2}$$


I'm not sure what to do after this step. The final answer is $$y = -x + 5$$
Thanks.

 

[arildno]

Your last line is wrong (remember to change + to -!).
We have:
$$\lim_{h\to{0}}(\frac{2+h}{h(1+h)}-\frac{2}{h})=\lim_{h\to{0}}\frac{2+h-2(1+h)}{h(1+h)}=\lim_{h\to{0}}\frac{-h}{h(1+h)}=\lim_{h\to{0}}\frac{-1}{(1+h)}=-1$$