# Help with Calc problem, limits

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:

OlderDan
Homework Helper
-x^3 + 2x - 1 = x - x^3 + x - 1 = x(1 - x^2) + x - 1

continue

working...

hk

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

OlderDan
Homework Helper
h_k331 said:
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
The numerator has what you need. I just gave you one step in the right direction. Look for another step.

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

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

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

Daniel.

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

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

Daniel.

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

hk

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