What is the limit of #( x^3 - 8 )/ (x-2)# as x approaches 2?

Question

63486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-17T01:17:33

The limit is #12.#

Notice how you have a difference of two cubes.

#x^3 - y^3= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3#

#=(x-y)(x^2 + xy + y^2)#

In our expression, we have #y = 2#:

#((x)^3 - (2)^3)/(x-2) = (cancel((x-2))(x^2 + 2x + 4))/cancel(x-2) = x^2 + 2x + 4#

At this point, the limit can be evaluated:

#=> color(blue)(lim_(x->2)(x^3 - 8)/(x-2))#

#= lim_(x->2) x^2 + 2x + 4#

#= (2)^2+2(2)+4#

#= 4 + 4 + 4#

#= color(blue)(12)#

Answer 2 · 2017-09-24T04:21:58

# 12.#

We can have another soln., if we use the following useful Standard Limit :

#lim_(x to a)(x^n-a^n)/(x-a)=n*a^(n-1)..............(star).#

Accordingly,

#lim_(x to 2)(x^3-8)/(x-2),#

#=lim_(x to 2)(x^3-2^3)/(x-2),#

#=3*2^(3-1)..........................................................[because, (star)],#

#=3*2^2,#

#=12,# as Respected EEt-AP has already obtained!