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Evaluating Limits using L'Hospital (2)

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 \(\displaystyle \frac{0}{6} = 0\)
 

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.
 

shamieh

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

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. :D
 

shamieh

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775