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!

Limit of a Function with Radicals in the Numerator

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Limit as h approaches 0 for [rad(5+h)-rad(5-h)]/h

    2. Relevant equations



    3. The attempt at a solution

    limit as h approaches 0 for [(5+h)-(5-h)]/h[rad(5+h)+rad(5-h)]

    limit as h approaches 0 for 2h/h[rad(5+h)+rad(5-h)]

    limit as h approaches 0 for h/[rad(5+h)+rad(5-h)]

    This was as far as I could get. Sorry if it's a little messy.
     
  2. jcsd
  3. Sep 6, 2011 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Limit as h approaches 0 for [rad(5+h)-rad(5-h)]/h

    That is Lim_(h to 0) [itex]\frac{\sqrt{5+h} - \sqrt{5-h}}{h} [/itex]

    Multiply numerator and denominator by [itex]\sqrt{5+h} + \sqrt{5-h} [/itex]
    When you work through steps, you obtain expression,...
    [itex]\frac{2}{\sqrt{5+h} + \sqrt{5-h}} [/itex]

    As h approaches 0, the expression approaches [itex]\frac{2}{\sqrt{5} + \sqrt{5}} [/itex]

    Simplifying to [itex] \frac{\sqrt{5}}{5}[/itex]. DONE.

    Note minor TEX/LATEX learning problems, "Lim as h approaches 0"
     
  4. Sep 6, 2011 #3
    Thank you for clearing that up. Instead of cancelling the 'h', I cancelled the 2 by mistake.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit of a Function with Radicals in the Numerator
  1. 2 radicals in a limit? (Replies: 11)

Loading...