Dec 9, 2013 Thread starter #1 S shamieh Active member Sep 13, 2013 539 lim x--> 1 \(\displaystyle \frac{x^4 - 3x^3 + 3x^2 - x}{x^4 - 2x^3 + 2x - 1}\) I got \(\displaystyle \frac{0}{6} = 0\)

lim x--> 1 \(\displaystyle \frac{x^4 - 3x^3 + 3x^2 - x}{x^4 - 2x^3 + 2x - 1}\) I got \(\displaystyle \frac{0}{6} = 0\)

Dec 9, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 That's incorrect. Since you showed no work, I can't tell you where you went wrong. I will tell you I applied L'Hôpital's Rule 3 times to get the value of the limit.

That's incorrect. Since you showed no work, I can't tell you where you went wrong. I will tell you I applied L'Hôpital's Rule 3 times to get the value of the limit.

Dec 9, 2013 Thread starter #3 S shamieh Active member Sep 13, 2013 539 re did the problem, didn;t take the deriv properly in the denom. Did you get \(\displaystyle \frac{-1}{4}\)

re did the problem, didn;t take the deriv properly in the denom. Did you get \(\displaystyle \frac{-1}{4}\)

Dec 9, 2013 Admin #4 M MarkFL Administrator Staff member Feb 24, 2012 13,775 No, that's no what I got either (and I verified my result using a CAS). If you show your work, I can address where you are going wrong.

No, that's no what I got either (and I verified my result using a CAS). If you show your work, I can address where you are going wrong.

Dec 9, 2013 Thread starter #5 S shamieh Active member Sep 13, 2013 539 wow I'm a idiot. I put 24 - 12 = 24.... -_-... I got \(\displaystyle \frac{1}{2}\) .. correct?

Dec 9, 2013 Admin #6 M MarkFL Administrator Staff member Feb 24, 2012 13,775 shamieh said: wow I'm a idiot. I put 24 - 12 = 24.... -_-... I got \(\displaystyle \frac{1}{2}\) .. correct? Click to expand... Yes, that's correct.

shamieh said: wow I'm a idiot. I put 24 - 12 = 24.... -_-... I got \(\displaystyle \frac{1}{2}\) .. correct? Click to expand... Yes, that's correct.