Results 1 to 4 of 4

Math Help - limits

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    60

    limits

    lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}

    lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0

    lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}

    lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer

    lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})

    can i verify the solution..thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    [quote=ashes;127090] lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}

    lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0

    lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}

    lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer

    lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})

    can i verify the solution..thanks...for the first one because of l'hopitals we have \lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}...applying L'hopitals rule again we get \lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)} splitting we get \lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2...and finally differntiatting on the first one we et \lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    [quote=Mathstud28;127109]
    Quote Originally Posted by ashes View Post
    lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}

    lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0

    lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}

    lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer

    lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})

    can i verify the solution..thanks...for the first one because of l'hopitals we have \lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}...applying L'hopitals rule again we get \lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)} splitting we get \lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2...and finally differntiatting on the first one we et \lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1

    The second and third one are just more simple applications of L'hopitals if I have time tonight I will show them...here is the fourth one \lim_{r \to 1}\frac{ar^{n}-a}{r-1}....since it is \frac{0}{0}...we can apply L'hopital's rule to get \lim_{r \to 1}\frac{anr^{n-1}}{1}=an
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    And I will give you a hint for the last one

    Quote Originally Posted by ashes View Post
    lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}

    lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0

    lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}

    lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer

    lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})

    can i verify the solution..thanks
    if you combine them you have \lim_{x \to 0}\frac{x-sin(x)}{xsin(x)}....hmm....I wonder what you could use on that...*cough*L'hopital's*cough*
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 05:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 23rd 2008, 11:17 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum