limit/l'hopital

**hazecraze** · November 13th 2009, 11:47 AM

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

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

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

**Deadstar** · November 13th 2009, 11:55 AM

The derivative of the top line should be...
$3ax^{3a-1} - 3a$

**hazecraze** · November 13th 2009, 03:30 PM

so $f'$ of numerator and denominator $= \frac{(3a)(3a -1)[(3ax)^(3a-2)]}{2}$ ?

**hazecraze** · November 14th 2009, 05:46 PM

I simplified it to this,:

but no luck.

**Prove It** · November 14th 2009, 05:51 PM

Originally Posted by hazecraze

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

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

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

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

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

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

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

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

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