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

1 Answer
Jun 16, 2018

#(5alpha^3-15alpha^2+1)/(3alpha^5-12alpha+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
#(5alpha^3-15alpha^2+1)/(3alpha^5-12alpha+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.