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

    AKG

    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

    HallsofIvy

    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|)
    x->-2

    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