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

Answer 1

Isabella Ackerman

#16/3#

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Answer 2

Christopher Allers

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

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#

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Answer 3

Isaiah Allen

#16/3#

#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#

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Answer 4

William Abdalla

The limit of (x^4 - 256) / (x^3 - 64) as x approaches 4 is 48.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.