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 as h->0 when [cos(pi/2+h)]/h

  1. Oct 22, 2004 #1
    limit as h->0 when [cos(pi/2+h)]/h

    when I sub in zero...I get 0/0...so I have to use L'hopital's rule
    So I take the derivative of the numberator and denominator...

    =[-sin(pi/2+h)*(0+1)]/1
    =-sin(pi/2+h)
    =-sin(pi/2+0)
    = -1

    is this correct?
     
  2. jcsd
  3. Oct 22, 2004 #2
    First off do you mean

    [tex]\cos(\frac{\pi}{2} + h)[/tex]
    or
    [tex]\cos(\frac{\pi + h}{2})[/tex]

    I think you mean the first expression

    Note that [tex]\cos(\frac{\pi}{2} + h) = -\sin(h)[/tex]
    and the given limit equals
    [tex]-\lim_{h->0} \frac{\sin h}{h} = -1[/tex]

    At this juncture you may use L Hospital's Theorem to reduce it to cos(h)/h but the fraction sin(h)/h goes to 1 in the limit as can be proved without L-Hospital's Theorem.

    Your solution is correct.

    Cheers
    Vivek
     
    Last edited: Oct 22, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit as h->0 when [cos(pi/2+h)]/h
  1. H=rt-4.9t^2 questions (Replies: 17)

  2. MSv/h to R/h (Replies: 6)

Loading...