What is the limit of ( x^3 - 8 )/ (x-2)x3−8x−2 as x approaches 2?
2 Answers
The limit is
Explanation:
Notice how you have a difference of two cubes.
x^3 - y^3= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3x3−y3=x3+x2y+xy2−x2y−xy2−y3
=(x-y)(x^2 + xy + y^2)=(x−y)(x2+xy+y2)
In our expression, we have
((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)
Explanation:
We can have another soln., if we use the following useful Standard Limit :
Accordingly,
