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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook