Results 1 to 4 of 4

Thread: [SOLVED] Limit question

  1. #1
    PTL
    PTL is offline
    Junior Member
    Joined
    Aug 2009
    Posts
    29

    [SOLVED] Limit question

    limit as h goes to 0 of$\displaystyle \frac{f(2+h)-f(2)}{h}$ where f(x) = $\displaystyle \sqrt{x}$

    It looks similar to differentiation from first principles, but I can't see how to use that...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Remember that $\displaystyle (a-b)(a+b)=a^2-b^2$

    This is a common trick : multiply by $\displaystyle 1=\frac{f(2+h)+f(2)}{f(2+h)+f(2)}$

    The numerator will simplify with the denominator, and what's left is no longer undefined
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    612
    Thanks
    309
    Quote Originally Posted by PTL View Post
    limit as h goes to 0 of$\displaystyle \frac{f(2+h)-f(2)}{h}$ where f(x) = $\displaystyle \sqrt{x}$

    It looks similar to differentiation from first principles, but I can't see how to use that...
    $\displaystyle \lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h}$

    Rationalize the numerator:
    $\displaystyle = \lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h} \cdot \frac{\sqrt{2 + h} + \sqrt{2}}{\sqrt{2 + h} + \sqrt{2}}$

    $\displaystyle = \lim_{h \to 0} \frac{2 + h - 2}{h(\sqrt{2 + h} + \sqrt{2})}$

    $\displaystyle = \lim_{h \to 0} \frac{h}{h(\sqrt{2 + h} + \sqrt{2})}$

    $\displaystyle = \lim_{h \to 0} \frac{1}{\sqrt{2 + h} + \sqrt{2}}$

    Now you can evaluate the limit by direct substitution.


    01


    EDIT: too slow...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by PTL View Post
    limit as h goes to 0 of$\displaystyle \frac{f(2+h)-f(2)}{h}$ where f(x) = $\displaystyle \sqrt{x}$

    It looks similar to differentiation from first principles, but I can't see how to use that...
    If you are supposed to know that $\displaystyle f$ is differentiable at $\displaystyle x=2$ then you can claim that
    $\displaystyle \lim_{h\to0}\frac{f(2+h)-f(2)}{h}=f'(2)$.
    and all you have to do is compute $\displaystyle f'(2)$...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Limit and integral question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 9th 2009, 05:06 PM
  2. [SOLVED] infinite series limit question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 22nd 2009, 03:25 PM
  3. [SOLVED] another trig limit question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 19th 2009, 09:36 PM
  4. [SOLVED] Trig limit Question
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Sep 26th 2009, 09:13 PM
  5. Replies: 10
    Last Post: Jul 20th 2008, 01:17 PM

Search Tags


/mathhelpforum @mathhelpforum