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!

LIMIT question needs your attention

  1. Sep 15, 2011 #1
    Limit x-->0- [[sqrt(88*x^2)] / x]
    i think the answer of this question should be sqrt(88) but according to my instructor its -sqrt(88)
    so please solve this one for me. only the left hand limit .
     
  2. jcsd
  3. Sep 15, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Remember that sqrt(x2)= |x| = -x if x < 0.
     
  4. Sep 15, 2011 #3
    ok...so the x in numerator will be positive and the one in the denominator negative.
    so now the answer will be -sqrt(88)
     
  5. Sep 15, 2011 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not quite. The x in the numerator is equal to the x in the denominator; it's not that one of them is negative and the other positive. The point is that the fraction |x|/x = (-x)/x. Still, x itself is negative as x → 0-.
     
  6. Sep 15, 2011 #5
    i have one more question for you
    cant we cut the two 'x' in the numerator and denominator without putting values?
    if we do so then we wont get the - sign in the answer
     
  7. Sep 15, 2011 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can't cancel the x values in |x|/x without distinguishing the cases whether x > 0 or x < 0.
     
  8. Sep 15, 2011 #7
    sorry.i get your point now. but your last point has confused them in another problem:
    lim x->5 [3x-15/(sqrt{x^2 + 25 - 10x}]
    how will u solve this question for the left and right hand limits separately?
     
  9. Sep 15, 2011 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Given our previous example, what do you get for your denominator [itex]\sqrt{(x-5)^2}[/itex] ? It is the same idea.
     
  10. Sep 15, 2011 #9
    here is what i understand:
    for right hand limit : 3(x-5)/|x-5|
    and since both numerator and denominator are positive we get 3

    for left handed : 3(x-5)/|x-5|
    in this case the numerator becomes negative but the denominator stays positive . so we get -3
    ryt?
    and overall the limit of function doesnt exist at 3
     
  11. Sep 15, 2011 #10

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    :smile: Yes that's right. Sorry for the delay getting back to you but the rest of my life intervened.
     
  12. Sep 16, 2011 #11
    thanks for your help...you cleared my concept
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: LIMIT question needs your attention
Loading...