Results 1 to 8 of 8

Math Help - Limits with L'Hopital

  1. #1
    Member
    Joined
    Jul 2005
    Posts
    187

    Limits with L'Hopital

    Need help with these Hospital limits:
    Attached Thumbnails Attached Thumbnails Limits with L'Hopital-untitled.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Find: \lim_{x\to 0}\tan(x)\,\ln(x).

    Rewrite as:

    \lim_{x\to 0}\frac{\ln(x)}{\cot(x)},

    now use L'hopital's rule:

     \lim_{x\to 0}\frac{\ln(x)}{\cot(x)}=\lim_{x \to 0}\frac{1/x}{(-\csc^2(x))},

    ...............=  \lim_{x \to 0}-\frac{\sin^2(x)}{x}

    ...............= 0

    RonL
    Last edited by CaptainBlack; December 17th 2006 at 09:55 AM. Reason: to correct a typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Find \lim_{x \to 0}\frac{1}{x^2}-\frac{\sin(x)}{x}

    The first term in the limit clearly goes to \infty, while the second
    term is a well known limit equal to 1, so the limit as a whole is \infty.


    Opps didn't see the actual question:

    Find \lim_{x \to 0}\frac{1}{x^2}-\frac{\cos(x)}{\frac{\sin(x)}{x}}

    but the second term still has a finite limit and so the whle thing still goes to \infty

    (or is a different interpretation of the second term in the limit meant?)


    RonL
    Last edited by CaptainBlack; December 17th 2006 at 10:17 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    768
    Hello, totalnewbie!

    Here's the first one . . .


    1)\;\lim_{x\to0}\left(\tan x\cdot\ln x\right)

    The function goes to: (0)(-\infty)

    Rewrite as: . \lim_{x\to0}\frac{\ln x}{\cot x} . . . which goes to \frac{-\infty}{\infty}

    Apply L'Hopital: . \lim_{x\to0}\left(\frac{\frac{1}{x}}{-\csc^2x}\right)\;=\;\lim_{x\to0}\left(-\frac{\sin^2x}{x}\right)

    . . = \;\lim_{x\to0}\left[(-\sin x)\left(\frac{\sin x}{x}\right)\right] \;= \;-(0)(1) \;=\;\boxed{0}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2005
    Posts
    187
    I used Hospital rule for the second one too and got that final result is cosx*sinx. 0*1=0
    Just want to see someone's confirmation.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Third go at the second question:

    Find \lim_{x \to 0} \frac{1}{x^2}-\frac{\cos(x)/\sin(x)}{x}

    which seems a plausible form for what might have been intended.

    (the limit is 1/3 but I will need to do some work to produce what you would
    think an acceptable explanation)

    The limit is:

    \lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}

    Take the series expasion of \cot(x)=\frac{1}{x}-\frac{x}{3}+O(x^3)

    So:

    \lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}=\lim_{x \to 0}\frac{1}{x^2}-(1/x)\left[\frac{1}{x}-\frac{x}{3}+O(x^3)\right]

    .............. =\lim_{x \to 0}\frac{1}{3}-O(x^2)=\frac{1}{3}



    RonL
    Last edited by CaptainBlack; December 17th 2006 at 11:06 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    768
    Hello, totalnewbie!

    I am assuming that the second problem is: . \lim_{x\to0}\frac{1}{x^2} - \frac{\frac{\cos x}{\sin x}}{x}

    Then we have: . \frac{1}{x^2}-\frac{\cos x}{x\sin x} \;=\;\frac{\sin x - x\cos x}{x^2\sin x} . . . which goes to \frac{0}{0}


    Apply L'Hopital: . \frac{\cos x + x\sin x - \cos x}{x^2\cos x + 2x\sin x} \:=\:\frac{x\sin x}{x^2\cos x + 2x\sin x} = \:\frac{\sin x}{x\cos x + 2\sin x}


    Divide top and bottom by \sin x:\;\;\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}


    Then: . \lim_{x\to0}\left(\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}\right) \;=\;\frac{1}{1\!\cdot1 + 2} \;=\;\boxed{\frac{1}{3}}

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2005
    Posts
    187
    Thank you Soroban!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Limits using L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2011, 02:53 PM
  2. Two tricky L'Hopital limits
    Posted in the Calculus Forum
    Replies: 11
    Last Post: January 12th 2011, 01:12 PM
  3. L'Hopital/limits question
    Posted in the Calculus Forum
    Replies: 21
    Last Post: January 8th 2011, 01:15 PM
  4. Using L'Hopital's Rule To Evaluate Limits
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 8th 2010, 03:37 AM
  5. limits(l'hopital's rule)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2008, 06:42 AM

Search Tags


/mathhelpforum @mathhelpforum