Results 1 to 10 of 10

Math Help - Help with limits

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    30

    Help with limits

    I have a problem with my homework:
    1. prove only by the defenition of the limit that <br />
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{1}{{\cos x}} \ne \infty <br /> <br />

    Thanks ahead
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by omert View Post
    I have a problem with my homework:
    1. prove only by the defenition of the limit that <br />
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{1}{{\cos x}} \ne \infty <br /> <br />

    Thanks ahead
    Are you allowed to use L'Hospital's Rule?

    If so, it's very easy...

    \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos{x}} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1)}{\frac{d}{dx}(\cos{x})}

     = \lim_{x \to \frac{\pi}{2}} \frac{0}{-\sin{x}}

     = \frac{0}{-\sin{\frac{\pi}{2}}}

     = \frac{0}{-1}

     = 0.


    Without L'Hospital, not so easy...

    All I can think of is the Sandwich Theorem, but I can't think of what to sandwich it inside.

    I know that \frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}.

    We just need to find something that \frac{1}{\cos{x}} is less than or equal to...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    Are you allowed to use L'Hospital's Rule?

    If so, it's very easy...

    \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos{x}} = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1)}{\frac{d}{dx}(\cos{x})}

     = \lim_{x \to \frac{\pi}{2}} \frac{0}{-\sin{x}}

     = \frac{0}{-\sin{\frac{\pi}{2}}}

     = \frac{0}{-1}

     = 0.


    Without L'Hospital, not so easy...

    All I can think of is the Sandwich Theorem, but I can't think of what to sandwich it inside.

    I know that \frac{\sin{x}}{x}\leq \frac{1}{\cos{x}}.

    We just need to find something that \frac{1}{\cos{x}} is less than or equal to...
    The given limit is not of indeterminant form, l'Hopitals rule is not valid. The limit is not equal to zero.

    The limit does not exist. My suggestion is to consider the left hand and right hand limits ....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by mr fantastic View Post
    The given limit is not of indeterminant form, l'Hopitals rule is not valid. The limit is not equal to zero.

    The limit does not exist. My suggestion is to consider the left hand and right hand limits ....
    Oops, my mistake... I knew I was missing something obvious
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2008
    Posts
    30
    I can't use any theorem. I need to prove it only by the defenition. let be M>0 so the any delta>0.... x-pi/2 -->>> 1/cosx<=M
    thanks ahead
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by omert View Post
    I can't use any theorem. I need to prove it only by the defenition. let be M>0 so the any delta>0.... x-pi/2 -->>> 1/cosx<=M
    thanks ahead
    You could just look at the graph...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    You need to approach \pi/2 from the right and the left.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    Like matheagle says, approach from left and right.

     \lim_{x\rightarrow \frac{\pi}{2}} \frac{1}{cos x} = \infty

    And from the left:

     \lim_{x\rightarrow \frac{-\pi}{2}} \frac{1}{cos x} = -\infty

    I hope this is correct..
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    you meant....

    Quote Originally Posted by Twig View Post
    Like matheagle says, approach from left and right.

     \lim_{x\rightarrow \pi/2^+} \frac{1}{cos x} = -\infty

    And from the left:

     \lim_{x\rightarrow \pi/2^-} \frac{1}{cos x} = \infty
     -\pi/2 is not   \pi/2^-

    And the limits were backwards.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    oops yeah, thx
    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