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Help with Calc problem, limits

  • Thread starter h_k331
  • Start date
33
0
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

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

continue
 
33
0
working...

hk
 
33
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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

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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.
 
33
0
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

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[tex] -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] [/tex]

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

Daniel.
 
33
0
dextercioby said:
[tex] -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] [/tex]

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

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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
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:

[tex]\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[/tex]

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

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