1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Need a little help please.

  1. Jun 22, 2005 #1
    Hello everybody. I just need a little help with some very basic Calculus. Actually I need help with the Algebra part, but it is a Calculus problem. Here is the problem.
    Lim x->-2 of (x^2+3x+2)/(2-|x|)

    That is it and I know the answer is -1, but I cannot get that |x| out of the denomenator, I have tried to multiply by the conjugate, but that did not seem to work. Thanks in andave for any advice.
    Last edited: Jun 22, 2005
  2. jcsd
  3. Jun 22, 2005 #2
    Is it x^2+3x+2 or x^2+3x-2?
  4. Jun 22, 2005 #3
    He must mean
    [tex]\lim_{x\rightarrow -2}\frac{x^2+3x+2}{2-|x|}[/tex]
    Otherwise it's undefined.
  5. Jun 22, 2005 #4
    I edited my post and Vegeta was correct.
  6. Jun 22, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    As x approaches -2, x is negative.
  7. Jun 22, 2005 #6
    Think of the absolute value funciton as a piecewise function, defined separately for positive numbers and negative numbers.
  8. Jun 22, 2005 #7
    Yes and keep in my to factor x^2+3x+2
  9. Jun 22, 2005 #8
    I have thought of it as a piecewise function, but my instructions are to solve this algebraically. I have also factored it ((x+1)(x+2))/(2-|x|). Now what? Thanks for the help.
  10. Jun 22, 2005 #9
    Defining piecewise functions is an algebraic method.
  11. Jun 22, 2005 #10
    As you approach -2 from left and right, lxl is defined as -x. Try to use this.
  12. Jun 23, 2005 #11
    x is having the negative value in this case then IxI will have I-2I=-2
    so the 2-IXI term will be 4 simple
  13. Jun 23, 2005 #12


    User Avatar
    Science Advisor

    No, for x close to -2, the denominator will be 2-|x|= 2-(-x)= 2+x. That's what you need. Now, what is the limit?
  14. Jun 23, 2005 #13
    Yes, thank you guys for all the help. Now I see that as you are close to -2, |x| is defined as -x so the problem looks something like this.

    lim ((x+1)(x+2))/(2-|x|)

    since we said |x| is -x we get 2-(-x) which cancels with the numerator and we are left with (x+1) and after pluging in the limit we get the answer which is -1.

    Thanks for all the help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook