Results 1 to 12 of 12

Math Help - Limits and continuity...

  1. #1
    Member
    Joined
    May 2008
    Posts
    102
    Awards
    1

    Limits and continuity...

    Hello...

    I have got some HW problems based on limits and continuity...please help in solving those problems..

    Q1) Find the limit of the following using L'Hospital rule....

    a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

    b) lim x-> pi/2 [pi/2 - x] . tan(x)


    Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

    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 Originally Posted by Vedicmaths View Post
    Hello...

    I have got some HW problems based on limits and continuity...please help in solving those problems..

    Q1) Find the limit of the following using L'Hospital rule....

    a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

    b) lim x-> pi/2 [pi/2 - x] . tan(x)


    Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

    Thanks!
    Must you use L'hopital's?

    Ok for the first one seeing that if we let sin(x)-x=f(x) and g(x)=e^x-1

    and g(x)=f(x)=0

    Then \lim_{x\to{0}}\frac{f(x)}{g(x)}=\lim_{x\to{0}}\fra  c{f'(x)}{g'(x)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Melbourne
    Posts
    428
    Edit: duplicated mathstuds fine work, sorry

    L'Hopital's rule states that \lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)} if f(x_0)=0 and g(x_0) = 0 or f(x_0) = \pm \infty and g(x_0) = \pm \infty
    Go for it. Question 1 is easy.

    For question 2 start by taking the derivative of all parts of the inequality.
    Last edited by badgerigar; June 23rd 2008 at 06:46 PM. Reason: fixed latex error, then saw previous post
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Vedicmaths View Post
    Hello...

    I have got some HW problems based on limits and continuity...please help in solving those problems..

    Q1) Find the limit of the following using L'Hospital rule....

    a) lim x-> 0+ [sin(x) - x] / [e^x - 1]

    b) lim x-> pi/2 [pi/2 - x] . tan(x)


    Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1

    Thanks!
    The first one is of the form \frac{0}{0} so now using L'hospitals rule we take a derivative of the numerator and the denominator to get


    \lim_{x \to 0^+}\frac{\cos(x)-1}{e^x}=\frac{\cos(0)-1}{e^{0}}=\frac{1-1}{1}=0

    For the 2nd rewrite it as \frac{[x-\pi/2]\sin(x)}{\cos(x)}

    now as x\to \frac{\pi}{2} this is of the form \frac{0}{0}

    Using L.H again we get

    \lim_{x \to \frac{\pi}{2}}\frac{[x-\pi/2]\cos(x)+\sin(x)}{-\sin(x)}=\frac{[0]0+1}{-1}=-1

    For the 2nd question consider the function

    f(x)=\ln(x) f is continous on [1,\infty) and differentiable on (1,\infty) so it satisfies the hypothesis of the MVT

    by the MVT f(b)-f(a)=f'(c)(b-a) where c \in (a,b)

    Let a=1 and b=x and we get

    \ln(x)-\ln(1)=\frac{1}{c}\left( x-1 \right) simplifying we get

    \ln(x) =\frac{x-1}{c} \iff c\ln(x)=x-1

    Since c \in (1,x) \mbox{ or } 1< c < x
     \ln(x) < x-1

    Also by the same reasoning c < x \iff \frac{1}{x} < \frac{1}{c} so

    \frac{x-1}{x}< \frac{x-1}{c}=\ln(x)

    so finally we get

    \frac{x-1}{x} < \ln(x) < x-1

    Yeah!!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Is it necessary L'H˘pital?

    Q2 can also be killed as follows:

    Since \ln x=\int_{1}^{x}{\frac{dy}{y}}, we have \frac{1}{x}\le \frac{1}{y}\le 1\implies \frac{x-1}{x}\le \ln x\le x-1.\quad\blacksquare (This holds for x>0.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    102
    Awards
    1
    Thanks a lot!

    That really helps alot!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2008
    Posts
    102
    Awards
    1
    Dear Empty Set!

    In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Vedicmaths View Post
    Dear Empty Set!

    In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference?

    Thanks!
    \lim_{x \to \frac{\pi}{2}}\frac{[x-\frac{\pi}{2}]\sin(x)}{\cos(x)}=\frac{[\frac{\pi}{2}-\frac{\pi}{2}]\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{[0]\cdot 1}{0}=\frac{0}{0}

    So now we can use L'hospitals rule.

    Then we can take the derivatives to get what I did above.

    I hope this helps.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2008
    Posts
    102
    Awards
    1

    confusion!

    Yes Sir! I got what you were trying to do after using the L'hospital Rule but I was asking that why are you manipulating the term? Since, in the problem it was given as ( pi/2 - x) sinx / cosx...but you have mentioned that we can rewrite the function as

    So why are you changing the sign here? And I guess that will change the answer. I got 1 from the function given in the original question and you are getting -1 from this one.
    I am sorry to bother you. But I am just a little bit confused and want to learn if something I do not know.

    Thanks so much once again!

    Regards,
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    The original problem was this

    Quote Originally Posted by Vedicmaths View Post

    b) lim x-> pi/2 [pi/2 - x] . tan(x)
    If we take the limit as x \to \frac{\pi}{2}

    we get 0\cdot \pm \infty

    This doesnt quite fit the form to use L'hospitals rule.

    We need to manipulate the equation into the form \frac{0}{0} or \frac{\infty}{\infty}

    Hence I changed [\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}

    Now this is of the form \frac{0}{0}


    So from here we take the derivative of the numerator and the denomiator as per L.H's rule.

    I guess I don't understand what you mean changing the sign?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by TheEmptySet View Post
    The original problem was this



    If we take the limit as x \to \frac{\pi}{2}

    we get 0\cdot \pm \infty

    This doesnt quite fit the form to use L'hospitals rule.

    We need to manipulate the equation into the form \frac{0}{0} or \frac{\infty}{\infty}

    Hence I changed [\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}

    Now this is of the form \frac{0}{0}


    So from here we take the derivative of the numerator and the denomiator as per L.H's rule.

    I guess I don't understand what you mean changing the sign?
    You did
    Quote Originally Posted by TheEmptySet View Post
    For the 2nd rewrite it as \frac{[x-\pi/2]\sin(x)}{\cos(x)}
    Which is the negative of the original problem.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Sorry, I didn't even notice that I switched them.

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limits and continuity
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 3rd 2010, 01:58 PM
  2. [SOLVED] Continuity with limits
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 17th 2010, 07:45 AM
  3. Limits and Continuity Help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 19th 2009, 10:15 PM
  4. continuity and limits
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 28th 2008, 04:48 AM
  5. Limits and Continuity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 3rd 2007, 07:31 PM

Search Tags


/mathhelpforum @mathhelpforum