# Evaluate Lim 5x³-15x²+1/3x^5-12x+5? X=α

Jun 16, 2018

$\frac{5 {\alpha}^{3} - 15 {\alpha}^{2} + 1}{3 {\alpha}^{5} - 12 \alpha + 5}$

#### Explanation:

I wonder if I am misunderstanding something here. As far as I can see, the limit of 5x³-15x²+1/3x^5-12x+5 when $x = \alpha$ should simply be
1/3alpha^5+5alpha³-15alpha²-12alpha+5

If you actually mean
(5x³-15x²+1)/(3x^5-12x+5)
please remember to write it as (5x³-15x²+1)/(3x^5-12x+5), otherwise you confuse at least some people, including me.

I still get the answer
$\frac{5 {\alpha}^{3} - 15 {\alpha}^{2} + 1}{3 {\alpha}^{5} - 12 \alpha + 5}$
though, but the question is if it's possible to factorise the denominator and numerator, and if they have any factor in common. Offhand I cannot see it, as it needs some work.