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#