Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the Limit

  1. Dec 8, 2007 #1
    Hello there
    Find the Limit of this function without using the method mentioned below.

    The Cubic root of X - 4
    ____________________

    The Square root of X - 8


    X --> 64

    Find the Lim without using this method

    X^N - A^N
    _________ = N over M times A^(N-M)
    X^M - A^M
     
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Gib Z

    User Avatar
    Homework Helper

    Expand the powers of x with Newton's Generalization of the Binomial Theorem.
     
  4. Dec 8, 2007 #3
    Can you write the solution please?

    Thanks.
     
  5. Dec 8, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    No. Sorry.
     
  6. Dec 8, 2007 #5
    Thanks generally.
     
  7. Dec 8, 2007 #6
    Am I not getting this or what?

    Hi,

    I wanted to answer your question and it seems simple to me. The expression is:

    (x-4)^(1/3)) div by (x-8)^(1/3) and x approaches 64. Just substitute x= 64 and thats the answer right? The denominator does not go to zero! Maybe I have the expression wrong.

    Am I not getting this or what?
     
  8. Dec 8, 2007 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, you have the expression wrong. It is once again a lack of parentheses (and in the denominator, square root, not cube root.) In order that this problem NOT be trivial, the original post must have meant ( cube root(x)- 4)/(square root(x)- 8).

    [tex]\frac{^3\sqrt{x}- 4}{\sqrt{x}- 8}[/tex].
     
  9. Dec 8, 2007 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I am completely mystified! You say at the beginning, "using the method mentioned below" but at the end say "without using this method".

    And if you mean "using this method" I am still mystified because the equation you cite simply isn't true! The left side is a function of X and the right isn't!

    I'm inclined to think that you meant
    [tex]\frac{X^N- A^N}{X^M-A^M}= \frac{(X-A)(X^{N-1}+ AX^{N-2}+ \cdot\cdot\cdot+ A^{N-2}X+ A^{N-1})}{(X-A)(X^{M-1} AX^{M-2}\cdot\cdot\cdot+ A^{M-2}X+ A^{M-1})}[/tex]
    which works for M and N positive integers. Witn M= 1/3, and N= 1/2, that becomes "Newton's Generalization of the Binomial Theorem" that GibZ mentioned. But are you required to use it or not allowed to use it?
     
  10. Dec 8, 2007 #9
    The problem is precisely as you mentioned, that's what I meant and I am sorry about miswriting the problem because I wanted to write the problem fast without having to search for the codes to write it.

    The Newton's Generalization of the Binomial Theorem is not allowed, I just have known that this is called what you called it because my textbooks are not written in English.

    I tried to solve it many times but I usually end up with Zero whether it is numerator or denominator, there are many problems that usually get solved if you mulitpy it times the numerator or denominator and sometimes both but This problem is different.

    by the way, the answer by Newton's law is 1 over 6, there are two things, whether this problem is not solvable by Algebriac manipulation or it can only be solved by Graphing it.

    I guess this is a problem from a mathematics competition because the one who gave it to me got it from someone who participates in mathematics competitions who apparently got this problem from the math teacher.
     
    Last edited: Dec 8, 2007
  11. Dec 8, 2007 #10
    [tex]\frac{^3\sqrt{x}- 4}{\sqrt{x}- 8}[/tex]

    If you Plug X by 64, it is gonna be (4-4) over (8-8) which equals 0 over 0
     
  12. Dec 8, 2007 #11
    Which implies that the poster could try using L'Hôpital's rule.
     
  13. Dec 10, 2007 #12
    You can also use L'Hopitals rule here. Oh sorry, someone already suggested it. The answer I got is 1/3 and I am sure of it.
     
    Last edited: Dec 10, 2007
  14. Dec 11, 2007 #13

    rbj

    User Avatar

    listen, racer, you need to express this in a manner that we can read it without digging out the secret decoder ring. try to use [tex]\LaTeX[/tex] a little.
     
  15. Dec 11, 2007 #14

    Gib Z

    User Avatar
    Homework Helper

    I know this isn't the homework forums, but this does indeed look like a homework question, so please do not give out answers, that in generally not appreciated. To the OP- I did the working and I get a different answer, though I could be wrong. You should check it yourself.
     
  16. Dec 11, 2007 #15
    Sorry Gib Z. I am new to the forum and didn't know. I will refrain from this in the future.
     
  17. Dec 11, 2007 #16
    Yeah that's the right answer.


    I know that there is a forum for homework helps, I would've posted it in the homework forum if it was a homework, actually it is final exams time.

    I don't need the solution but for everyone who is interested here, try doing it without Newton's method nor L'Hôpital's rule, instead, solve it by algebra if you can, don't write solution. just try do it yourself.

    Thanks.
     
    Last edited: Dec 11, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find the Limit
  1. Find the limit (Replies: 10)

  2. Finding limit (Replies: 1)

  3. Find the limit (Replies: 3)

  4. Finding the Limit (Replies: 1)

  5. Finding Limit (Replies: 11)

Loading...