1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about factoring for calculus

  1. Sep 10, 2006 #1
    Hello, I have been reading these forums for quite a while now and am very impressed! I have a question :) I need to find the lim(x->4) (k^2-16)/(sqrt(k)-2) How would I go about factoring k^2-16 so that I can get sqrt(k)-2 on the numerator? Thanks!
    Last edited: Sep 11, 2006
  2. jcsd
  3. Sep 11, 2006 #2
    [tex]\left(\sqrt{k}\right)^4 = k^2[/tex] and [tex]2^4 = 16[/tex].
  4. Sep 11, 2006 #3
    i would multiply the fraction by 1, where 1 = (sqrt(x) + 2)/(sqrt(x) + 2)

    i guess that's the 1st step & the only tricky part. isn't this in your textbook?
  5. Sep 11, 2006 #4
    Thankyou both. No unfortunately it is not in my textbook. I am using my father's calculus book from university as a supplement to my regular calculus material. It was probably assumed that I knew how to solve it so it wasn't even mentioned. Unfortunately I forgot how to solve it! Anyways, thanks.
  6. Sep 11, 2006 #5
    something you might have noticed is that it would be a very BAD idea to multiply the sqrt(x)+2 into the numerator, and probably not a good idea in general to do that. factor the x^4 - 16 & only multiply the denominators to get what you want to cancel.
  7. Sep 11, 2006 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    At some point prior to taking calculus, you should have learned the general formula (a2- b2)= (a- b)(a+ b).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?