# Limits & Derivatives

1. Oct 10, 2007

### Ydoc

1. The problem statement, all variables and given/known data
Find an equation of the tangent line to the curve at the given point.

y=(x-1)/(x-2), (3,2)

2. Relevant equations
3. The attempt at a solution
I've had several attempts at this, all of them are wrong. My solution book says the answer should be y=-x+5

lim as h->0 [(3+h-1)/(3+h-2) - (x-1)/(x-2)]/h

lim as h->0 [(h+2)/(h+1) - (x-1)/(x-2)]/h

and I lost track of where I wrote the next step to this attempt. I've already attempted this 4 times.

2. Oct 10, 2007

### bob1182006

you just need to use the general definition of a line:

(y-p)=m(x-r)

where (p,r) is the given point. and m is the slope.

To find the slope do you need to find the derivative and then evaluate for x=3.

But you don't need to use the definition of a derivative. just take the derivative of (x-1)/(x-2) using the quotient rule and then plug-in x=3 and you got your slope and plug in the point and slope into the formula and you'll get y=-x+5

3. Oct 10, 2007

### Ydoc

Thanks for the help.

The only problem I'm having currently is finding the derivative of (x-1)/(x-2). At this point in time my class hasn't learned the quotient rule yet, I checked the index and we don't learn it until the next chapter.

The only formula's that we were given in this section of this chapter is:

The tangent line to the curve y=f(x) at the point P(a,f(a)) is the line through P with slope:
m=lim as h->0 [f(a+h)-f(a)]/h

and

m=lim as x->a [f(x)-f(a)]/(x-a)

There's also a formula for instantaneous rate of change, but I don't think I need to use that for this problem.

I was able to do other problems like this, but they were stuff like y=x^2 .

Last edited: Oct 10, 2007
4. Oct 10, 2007

### bob1182006

o i see, well then you do have to use one of those definitions of the slope

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

from here find a common denominator, to combine the fraction into 1 and you should get h to cancel.