Results 1 to 8 of 8
Like Tree2Thanks
  • 1 Post By Ivanator27
  • 1 Post By Moebius

Math Help - Limit Involving Squeeze Theorem

  1. #1
    Member
    Joined
    May 2011
    Posts
    178
    Thanks
    6

    Limit Involving Squeeze Theorem

    The problem I have is \lim_{{x}\to{0}}\frac{cosx - 1}{x^2}

    I know I have to use the squeeze theorem, by implementing the range of the cosine function; but when I do that I get

    \frac{-1}{x^2}\leq\frac{cosx}{x^2}\leq\frac{1}{x^2}

    But that would imply the original limit would go to approach zero, but that is certainly not what the answer key attests. Am I setting up my range wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2012
    From
    South Africa
    Posts
    42
    Thanks
    3

    Re: Limit Involving Squeeze Theorem

    Do you have to use squeeze theorem? What about L'Opital's rule?
    Thanks from Bashyboy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2012
    From
    Arkansas
    Posts
    17
    Thanks
    1

    Re: Limit Involving Squeeze Theorem

    Hello Bashyboy,

    If we were to try to solve this problem with the squeeze theorem using the bound conditions for  \cos x , we would have:

    -1 \leq \cos x \leq 1

     \Rightarrow -2 \leq \cos x -1 \leq 0

     \Rightarrow \frac{-2}{x^2} \leq \frac{\cos x -1}{x^2} \leq \frac{1}{x^2}

     \Rightarrow -\infty \leq \lim_{{x}\to{0}} \frac{\cos x -1}{x^2} \leq \infty

    which doesn't really help us at all.


    Provided you are allowed to use L'Hospital's Rule, we can say:

     \lim_{{x}\to{a}} \frac{f(x)}{g(x)} = \lim_{{x}\to{a}} \frac{f'(x)}{g'(x)}

     \frac{d(\cos x -1)}{dx} = -\sin x

    and...

     \frac{d(x^2)}{dx} = 2x

    so...

     \lim_{{x}\to{0}} \frac{\cos x -1}{x^2} = \lim_{{x}\to{0}} \frac{-\sin x}{2x} = -1/2


    If you HAVE to use the squeeze theorem, just play around with functions that are ALWAYS outside of the bounds of  \cos x and see what happens. Good luck!
    Thanks from Bashyboy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2011
    Posts
    178
    Thanks
    6

    Re: Limit Involving Squeeze Theorem

    Ah, yes, that would be profoundly simple. Though, I wonder, would it be possible to do the squeeze theorem?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2012
    From
    Arkansas
    Posts
    17
    Thanks
    1

    Re: Limit Involving Squeeze Theorem

    I'm not really sure if you can do the squeeze theorem here. If it is possible, I definitely can't think of the bound functions at the moment lol.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,741
    Thanks
    481

    Re: Limit Involving Squeeze Theorem

    not a squeeze, but not L'Hopital either ...

    \frac{\cos{x}-1}{x^2} \cdot \frac{\cos{x}+1}{\cos{x}+1} =

    \frac{\cos^2{x}-1}{x^2(\cos{x}+1)} =

    \frac{-\sin^2{x}}{x^2(\cos{x}+1)}

    -\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}+1}

    now take the limit as x \to 0 ...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2012
    From
    london
    Posts
    10

    Re: Limit Involving Squeeze Theorem

    How do you find limit of sinx/x without L'hospital? That's not easier than original

    Quote Originally Posted by skeeter View Post
    not a squeeze, but not L'Hopital either ...

    \frac{\cos{x}-1}{x^2} \cdot \frac{\cos{x}+1}{\cos{x}+1} =

    \frac{\cos^2{x}-1}{x^2(\cos{x}+1)} =

    \frac{-\sin^2{x}}{x^2(\cos{x}+1)}

    -\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}+1}

    now take the limit as x \to 0 ...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,741
    Thanks
    481

    Re: Limit Involving Squeeze Theorem

    Quote Originally Posted by httr View Post
    How do you find limit of sinx/x without L'hospital? That's not easier than original
    at the start of most calculus courses, one learns the geometric proof of

    \lim_{x \to 0} \frac{\sin{x}}{x} = 1

    this basic trig limit is required in finding the derivative of y = \sin{x} using first principles ...

    f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin{x}}{h}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. squeeze theorem limit
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 1st 2012, 05:04 PM
  2. Evaluate the limit using the squeeze theorem?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 17th 2009, 12:03 PM
  3. Squeeze theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 8th 2009, 07:43 PM
  4. Need help with example of Squeeze Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 26th 2009, 04:03 AM
  5. Using the Squeeze Theorem to find a limit.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 27th 2008, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum