Homework Help: Help with Calc problem, limits

1. Jun 13, 2005

h_k331

This shouldn't be too hard but I'm stuck, brain is getting fried.

lim [(1+2x-x^3)-(1+2(1)-(1)^3)]/(x-1)
x->1

have gotten this far and I'm stuck:

lim (-x^3+2x-1)/(x-1)
x->1

Any help is appreciated, thx.

hk

ETA: fixed typo

Last edited: Jun 13, 2005
2. Jun 13, 2005

OlderDan

-x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1

continue

3. Jun 13, 2005

h_k331

working...

hk

4. Jun 13, 2005

h_k331

I really really want to just cancel x-1 from top and bottom! But knowing I can't, do I need to multiply top and bottom by something to change the denominator?

hk

5. Jun 13, 2005

OlderDan

The numerator has what you need. I just gave you one step in the right direction. Look for another step.

6. Jun 13, 2005

h_k331

Well, I gave up and solved it with the [f(a+h)-f(a)]/h formula instead of
[f(x)-f(a)]/(x-a). I was able to get the correct solution of -1 (according to the back of the book), but I'm still curious where I was going wrong with my algebra on the first try. Thanks for your help Dan, I appreciate it.

hk

7. Jun 14, 2005

dextercioby

$$-x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1=x(1+x)(1-x)-(1-x)=(1-x)[x(1+x)-1]$$

And you can see that the (1-x) has been factored.Simplify with the denominator and then take the limit.

Daniel.

8. Jun 14, 2005

h_k331

Doh! Difference of two squares, I tend to miss a lot of those when one of the squares is 1. Thanks Daniel.

hk

9. Jun 14, 2005

saltydog

I just wish to clear up something for the record:

When taking the limit, we are considering values of x close to 1 but NOT equal to 1. Thus it is possible to divide by (x-1). Not trying to upstage anyone here. Anyway, by synthetic division:

$$\mathop\lim\limits_{x\to 1}\frac{-(x^3-2x+1)}{x-1}=\mathop\lim\limits_{x\to 1}\frac{-(x^2+x-1)(x-1)}{x-1}=\mathop\lim\limits_{x\to 1}-(x^2+x-1)=-1$$

I know, I'm such a Johnny-come-lately.