Question

Question #c991c

Answer 1

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.

Answer 2

`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`