lim ( x^3 (x-3)^5 -64)/(x-4)
x→4
$\displaystyle
lim\frac{x^3(x-3)^5-64}{x-4}
$
use l'Hôpital's rule which says $\displaystyle lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}$
$\displaystyle
lim\frac{3x^2(5)(x-3)^4}{1}
$
fill in x as 4
$\displaystyle
lim\frac{3(4)^2(5)(4-3)^4}{1}
$
$\displaystyle
3(16)(5)(1^4)
$
$\displaystyle
lim=240
$
i think thats right..hope i helped
i think this is a easy but lengthy question. expand the numerator in the form (x^2-a^2)=(x-a)(x+a) as 64=2^3.after some calculations u will find that x-4 appears in the numerator as well. so x-4 gets cancelled. now put the linits and u get the required answer. if the calculations get too lengthy use L'Hospital's rule.
Why? L'Hopital's rule is by far the simplest way to do this. If you really refuse to use L'Hopital's rule, then you can multiply out $\displaystyle (x-3)^5$ to get the numerator is "standard form" then factor out (x-4) (since the numerator is 0 when x= 4, we know it must have such a factor)as Pulock2009 suggested.