how do i find the limits of X^3 - 27/ x- 3 as x approaches 3 ?

Answer 1

I believe the answer listed above me is incorrect. By asking as the limit x-3 approaches zero, it 'throws out' the concept of having the denominator equalling zero as 'bad'

Take x-3 and divide it into x^3-27 and you get x^2+3x+9
Just like 4 divided into 8 equals 2.

If you plug in 3 into x^2+3x+9 you get 9+9+9=27.

Answer is 27.

Answer 2

First rewrite the numerator as (x-3)(x^2+3x+9).
Then cancelling the common factors between the numerator and denominator, u have,
lim(x^2+3x+9) as x->3.
Now, apply the limit to (x^2+3x+9),
=3^2+3(3)+9 =9+9+9=27

Answer 3

Let y= lim {x^3 - 27} / (x-3)
x -->3

By using synthetic division,

x^3 - 27 = (x-3)(x^2 +3x + 9)

However, if you alert enough you will realise that (x^2 + 3x + 9) cannot be factorised as the Discriminant (b^2 -4ac) < 0 and also implies that it does not have any real roots.

Back to the question

=lim [(x-3)(x^2+3x+9)] / (x-3)
x -->3

=lim (x^2 + 3x + 9)
x -->3

= (3)^2 + 3(3) + 9

= 27

Answer 4

If you do the limit, you`ll get 0/0, which is impossible.
So, you must find a way to simplify.
If you notice, x^3 - 27 can also be written as
(x+3)(x+3)(x-3)

so now you can simplify the fraction and get
(x+3)(x+3)

So, your limit will be 9*9 =81

Answer 5

There are two ways to solve this problem.First one is by applying l,hospital rule which I think the easiest way to find the limit of the given problem. second one is the formal method.
Solution by applying l,Hospital rule:-
Given, X^3 - 27/ x- 3 as x approaches 3.
Let, f(x)= X^3 - 27; then, f '(x)=3x^2
And, let g(x)=x-3 ; then, g '(x)= 1
By l,Hospital rule, limit as x-->3, f(x)/g(x) =limit as x-->3,
f '(x)/g '(x) =3x^2=27
Second method:-
(x^3-27)/(x-3) =[(x-3)(x^2+3x+9)]/(x-3)=x^2+3x+9
limit as x-->3, x^2+3x+9=27

Answer 6

15

Answer 7

f(X) = (X^3 - 27)/( x- 3) =
(X^2 +3X + 3)( X - 3)/( X - 3) =
X^2 +3X + 3^2
When X is extreme small, the sum is about 27.
lim (for X--> 3) f(X) = f(3) = 27
The graph shows an extemely small gap at (3 , 27).
Nevertheless the function seems smooth around X=3.

Answer 8

Limit is 0~infinity

Answer 9

just sinply substitute 3 instade of x u'll get ans as15