Essentials of Calculus: find the limit problem

  • Thread starter Thread starter Nawz
  • Start date Start date
  • Tags Tags
    Calculus Limit
Click For Summary

Homework Help Overview

The problem involves finding the limit of the expression (x^3-27)/(x-3) as x approaches 3, which falls under the subject area of calculus, specifically limits and continuity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate the limit using a calculator but encounters confusion regarding the expression's setup. They consider factoring and substituting values but express uncertainty about the process. Other participants suggest factoring the expression and reference the identity for the difference of cubes. Some also mention alternative methods such as L'Hôpital's rule and substitution.

Discussion Status

The discussion is ongoing, with participants providing various approaches to tackle the limit problem. Suggestions include factoring the polynomial, using L'Hôpital's rule, and considering substitutions. There is no explicit consensus on the preferred method yet, as multiple strategies are being explored.

Contextual Notes

The original poster notes that plugging in the value 3 results in an indeterminate form (0/0), which raises questions about the assumptions made in evaluating the limit.

Nawz
Messages
32
Reaction score
0

Homework Statement



SYNTHESIS

Find limit as x approaches 3 of : (x^3-27)/x-3

Homework Equations





The Attempt at a Solution



I typed it into my calculator and got x^3 - (27)/(x-3) where x^3 wasn't divided by x-3. I really do not know where to start. I know the answer is 27 from the back of the book.

I was thinking it was something like x^3- (3)^3/ (x-3) and then you can cancel out one of the x-3 but I still cannot get 27 and if you plugin 3 into the original equation you get 0.
 
Last edited:
Physics news on Phys.org
Try factoring x3-27
 
You need to know that [itex](a- b)(a^2- ab+ b^2)= a^3- b^3[/itex].
 
Or, if you ever forget that a³ - b³ = (whatever it actually equals :-p )
It looks like something that can be factored.

Set x³ - 27 equal to zero.

x³ - 27 = 0

Find an x value that makes x³ - 27 = 0 hold, i.e. x = 3

(3)³ - 27 = 0

Then, because x = 3 is an answer, x - 3 = 0 is a factor so divide x - 3 into x³ - 27

______x²_+_..._______​
x - 3 |x³ + 0x² + 0x - 27
x³ - 3x²​
...​

Keep going & you'll factor it & get your answer :wink:
 
Or failing that L'Hopital's rule...

Mat
 
Or use a substitution with u = x - 3, no factoring cubes or l'Hôpital's rule. :wink:
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
24
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K