Struggling to Solve a Limit: \lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})

  • Context: Undergrad 
  • Thread starter Thread starter DDarthVader
  • Start date Start date
  • Tags Tags
    Difficulty Limit
Click For Summary
SUMMARY

The limit \(\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})\) can be solved by recognizing that \(x^3 + 1\) factors into \((x+1)(x^2-x+1)\). As \(x\) approaches -1, the expression results in an indeterminate form \(0/0\). To resolve this, one should simplify the expression using the factorization provided, leading to a clearer evaluation of the limit. The correct approach involves substituting the limit of the new variable \(u\) while addressing the indeterminate form.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with algebraic factorization
  • Knowledge of indeterminate forms
  • Experience with cube roots and their properties
NEXT STEPS
  • Study the concept of indeterminate forms in calculus
  • Learn how to factor polynomials, specifically cubic expressions
  • Explore techniques for evaluating limits, including L'Hôpital's Rule
  • Practice solving limits involving cube roots and rational functions
USEFUL FOR

Students studying calculus, particularly those focusing on limits and indeterminate forms, as well as educators seeking to clarify these concepts in their teaching.

DDarthVader
Messages
51
Reaction score
0
Hello! I'm having difficulty to find this limit: [tex]\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})[/tex]
This is what I'm trying to do to solve this limit: Let [tex]\frac{x^3+1}{x+1}=u[/tex] then
[tex]\lim_{x\rightarrow -1}(\sqrt[3]{u}) = \lim_{x\rightarrow -1}(\sqrt[3]{-1})[/tex]

I know something is wrong I'm just not sure about what is actually wrong. I'm thinking that [tex]\lim_{x\rightarrow -1}(\sqrt[3]{u})[/tex] Is wrong because that -1!

Thanks!
 
Physics news on Phys.org
DDarthVader said:
Hello! I'm having difficulty to find this limit: [tex]\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})[/tex]
This is what I'm trying to do to solve this limit: Let [tex]\frac{x^3+1}{x+1}=u[/tex] then
[tex]\lim_{x\rightarrow -1}(\sqrt[3]{u}) = \lim_{x\rightarrow -1}(\sqrt[3]{-1})[/tex]

I know something is wrong I'm just not sure about what is actually wrong. I'm thinking that [tex]\lim_{x\rightarrow -1}(\sqrt[3]{u})[/tex] Is wrong because that -1!

Thanks!



Hint: [itex]x^3+1=(x+1)(x^2-x+1)[/itex] .

DonAntonio
 
Also when you use the algebraic substitution, you need to consider the limit of the new variable u. So, as x--> -1 , what is the limiting value of u? However, this line of thiking will not save you from facing the indeterminate form (0/0). The easiest path is the hint that DonAntonio suggested.
 

Similar threads

  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K