Can Multivariable Limits Be Simplified Easily?

  • Context: MHB 
  • Thread starter Thread starter goody1
  • Start date Start date
  • Tags Tags
    Variables
Click For Summary
SUMMARY

The discussion focuses on simplifying multivariable limits, specifically the limit of the expression $\displaystyle \lim_{x \to a} \dfrac{(x^n-a^n) - na^{n-1}(x-a)}{(x-a)^2}$. Participants utilize L'Hôpital's Rule to evaluate the limit, ultimately arriving at the result $\dfrac{n(n-1)(a^{n-2})}{2}$. Key points include the clarification that only the variable x approaches a, while a and n remain constants, and the correction of misconceptions regarding the factorization of $x^n - a^n$.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of polynomial factorization
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the application of L'Hôpital's Rule in multivariable limits
  • Learn about polynomial factorization techniques
  • Explore advanced limit evaluation methods in calculus
  • Review the properties of continuous functions and their limits
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limit evaluation techniques, particularly in the context of multivariable functions.

goody1
Messages
16
Reaction score
0
Good evening! Could anybody help me with this limit?

limit.png

I have a problem when there are two variables. The only thing I did was that:

limit2.png
,
but I don't know if it was helpful. Thank you!
 
Physics news on Phys.org
What is the variable in the limit symbol?

Cbarker1
 
goody said:
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables. The only thing I did was that:

View attachment 10129,
but I don't know if it was helpful. Thank you!

actually, factoring out $(x-a)$ in the numerator yields ...

$\dfrac{(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ... + xa^{n-2} + a^{n-1} - na^{n-1})}{(x-a)^2}$

I played with the algebra to see if I could get the other $(x-a)$ factor in the denominator to divide out ... no joy, yet.

So, starting again ...

$\displaystyle \lim_{x \to a} \dfrac{(x^n-a^n) - na^{n-1}(x-a)}{(x-a)^2}$

using L'Hopital ...

$\displaystyle \lim_{x \to a} \dfrac{nx^{n-1} - na^{n-1}}{2(x-a)}$

$\displaystyle \lim_{x \to a} \dfrac{n(x^{n-1} - a^{n-1})}{2(x-a)}$

L'Hopital again ...

$\displaystyle \lim_{x \to a} \dfrac{n(n-1)(x^{n-2})}{2} = \dfrac{n(n-1)(a^{n-2})}{2}$
 
goody said:
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables.
No, there aren't. There are three symbols, x, a, and n. But only x is changing, it is going to a. a and n are constants. Only x is a variable.
The only thing I did was that:
View attachment 10129,
but I don't know if it was helpful. Thank you!
No. x^n- a^n is NOT (x- a)(x^{n-1}- a^{n-1}. If you multiply the right side of that you get x^n- a^{n-1}x^n- ax^{n-1}+ a^n. x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1}.
 
For Country Boy, "tex" tags don't work. Dollar signs and MATH tags still work.

Quoting the last part of your post ...

No. $x^n- a^n$ is NOT $(x- a)(x^{n-1}- a^{n-1})$.
If you multiply the right side of that you get $x^n- a^{n-1}x^n- ax^{n-1}+ a^n$.

$x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1})$.
 
Thank you. I am on entirely too many boards with too many different protocols!
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 53 ·
2
Replies
53
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K