Question #c991c

2 Answers
Oct 22, 2017

27

Explanation:

As it is hard to see that how to factorize the numerator and cancel out something with the denominator so as to make the denominator !=0, we can first let t^3=x and the lim will become:

lim_(x->27)(x-27)/(x^(1/3)-3)
=lim_(t->3 )(t^3-27)/(t-3)

In this form, we can factorize the denominator easier.
=lim_(t->3 ) [(t-3)(t^2+3t+3^2)]/(t-3)

=lim_(t->3)(t^2+3t+3^2)

=3^2+3*3+3^2 =27

It's the answer! Hope it can help u :)
Also, you can factorize the denominator using other way like the following:

lim_(x->27)(x-27)/(x^(1/3)-3)

lim_(x->27)([x^(1/3)]^3-3^3)/(x^(1/3)-3)

and further simplify it using a^3-b^3
But if you are not okay with this form of power of fraction, then you can try to following the first method.

Oct 22, 2017

27

Explanation:

(x-27)/(x^(1/3)-3) plugging in 27 gives the indeterminate form 0/0

Using l'Hospital's rule:

d/dx(x-27)=1

d/dx(x^(1/3)-3)=1/(3x^(2/3)

1/(1/(3x^(2/3)) )= 3x^(2/3)

Plugging in 27:

3(27)^(2/3)= 3root(3)(27^2)=3root(3)(729)=3*9=27

lim_(x->27)((x-27)/(x^(1/3)-3))=27