What is #lim_(xrarr4) ( x^4 -256) / (x^3 - 64)#?

3 Answers
Oct 27, 2015

#lim_(xrarr4)(x^4-256)/(x^3-64) = 16/3#

Explanation:

Either using polynomial long division or synthetic division we can discover:
#color(white)("XXX")(x^4-256) = (x-4)(x^3+4x^2+16x+64)#
and
#color(white)("XXX")(x^3-64)=(x-4)(x^2+4x+16)#

As long as #x!=4#
#color(white)("XXX")(x^4-256)/(x^3-64) = (x^3+4x^2+16x+64)/(x^2+4x+16)#

#lim_(xrarr4) (x^4-256)/(x^3-64) = ((4)^3+4(4)^2+16(4)+64)/((4)^2+4(4)+16)#

#color(white)("XXXXXXXXX")= (4(64))/(3(16)) = 16/3#

Oct 27, 2015

#16/3#

Explanation:

#lim_(x→4) (x^4 - 256)/(x^3-64)#

#=lim_(x→4) (x^4 - 4^4)/(x^3-4^3)#

i'm multiplying both numerator and denominator by #(x-4)#.
the reason for why i came up with this idea is that there is a theorem in limits as follows,
#lim_(x→a) (x^n - a^n)/(x-a) = n a^(n-1)#

so,
#=lim_(x→4) (x^4 - 4^4)/(x-4)*(x-4)/(x^3-4^3)#

#=lim_(x→4) (x^4 - 4^4)/(x-4)*lim_(x→4) (x-4)/(x^3-4^3)#

#= ( 4*4 ^ 3) /( 3 *4^2)#

#= 4^2/3#

#=16/3#

Oct 27, 2015

#16/3#

Explanation:

Arunraj