Is the Polynomial Limit Theorem Accurate and Comprehensive?

  • Context: Undergrad 
  • Thread starter Thread starter Orion1
  • Start date Start date
  • Tags Tags
    Limits Polynomial
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 14K views
Orion1
Messages
961
Reaction score
3


My Calculus professor has indicated a 'shortcut' in determining polynomial fraction limits, I am inquiring if this identity is correct, and how comprehensive is this 'theory'?

Polynomial Limit Theorem:
[tex]\lim_{x \rightarrow \infty} \frac{ax^2 - x + 2}{bx^2 - 1} = \frac{a}{b}[/tex]

[/Color]
 
Last edited:
Physics news on Phys.org
Orion1 said:
My Calculus professor has indicated a 'shortcut' in determining polynomial fraction limits, I am inquiring if this identity is correct, and how comprehensive is this 'theory'?
Polynomial Limit Theorem:
[tex]\lim_{x \rightarrow \infty} \frac{ax^2 - x + 2}{bx^2 - 1} = \frac{a}{b}[/tex]
[/Color]

Just divide through by the highest power, then the limit becomes A/B as x -> oo.
 
Last edited by a moderator:
You can also expand your Polynomial Limit Theorem like this:
Let [itex]m , \ n \in \mathbb{Z ^ +}[/itex]
If m < n:
[tex]\lim_{x \rightarrow \infty} \frac{\sum \limits_{i = 0} ^ m a_i x ^ i}{\sum \limits_{k = 0} ^ n b_k x ^ k} = 0 \quad (a_m, b_n \neq 0)[/tex]
If m > n:
[tex]\lim_{x \rightarrow \infty} \frac{\sum \limits_{i = 0} ^ m a_i x ^ i}{\sum \limits_{k = 0} ^ n b_k x ^ k} \quad (a_m, b_n \neq 0)[/tex] it does not have a limit.
If m = n:
[tex]\lim_{x \rightarrow \infty} \frac{\sum \limits_{i = 0} ^ m a_i x ^ i}{\sum \limits_{k = 0} ^ n b_k x ^ k} = \frac{a_m}{b_n} \quad (a_m, b_n \neq 0)[/tex]
 
Last edited: