limit/l'hopital

• Nov 13th 2009, 11:47 AM
hazecraze
limit/l'hopital
http://www.webassign.net/cgi-bin/sym...20-%201%29%5E2

lim
x->1 $\displaystyle \frac{3ax-3a}{2(x-1)}$

^I think the error is here, because if I simplify it, there is no where to do l'hopital.
x->1 $\displaystyle \frac{3a(x-1)}{2(x-1)}$

x->1 $\displaystyle \frac{3a}{2}$
• Nov 13th 2009, 11:55 AM
The derivative of the top line should be...
$\displaystyle 3ax^{3a-1} - 3a$
• Nov 13th 2009, 03:30 PM
hazecraze
so $\displaystyle f'$ of numerator and denominator$\displaystyle = \frac{(3a)(3a -1)[(3ax)^(3a-2)]}{2}$ ?
• Nov 14th 2009, 05:46 PM
hazecraze
I simplified it to this,:http://www.webassign.net/cgi-bin/pal...E%3C%2Fmath%3E
but no luck.
• Nov 14th 2009, 05:51 PM
Prove It
Quote:

Originally Posted by hazecraze
http://www.webassign.net/cgi-bin/sym...20-%201%29%5E2

lim
x->1 $\displaystyle \frac{3ax-3a}{2(x-1)}$

^I think the error is here, because if I simplify it, there is no where to do l'hopital.
x->1 $\displaystyle \frac{3a(x-1)}{2(x-1)}$

x->1 $\displaystyle \frac{3a}{2}$

$\displaystyle \lim_{x \to 1}\frac{x^{3a} - 3ax + 3a - 1}{(x - 1)^2} = \lim_{x \to 1}\frac{3ax^{3a - 1} - 3a}{2(x - 1)}$.

This is still indeterminate, so use L'Hospital again...

$\displaystyle = \lim_{x \to 1}\frac{3a(3a - 1)x^{3a - 2}}{2}$

$\displaystyle = \frac{3a(3a - 1)}{2}$

$\displaystyle = \frac{9a^2 - 3a}{2}$.