Results 1 to 7 of 7

Thread: help with some problems?

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    17

    help with some problems?

    thanks ?
    Attached Thumbnails Attached Thumbnails help with some problems?-untitled-2.gif  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    We need to find the limits,
    $\displaystyle \lim_{x\to 0} x\sin \frac{1}{x}$
    We can write,
    $\displaystyle -1\leq \sin \frac{1}{x} \leq 1$
    Multiply by "x" (positive or negative equality is preseved),
    $\displaystyle -x\leq x\sin \frac{1}{x} \leq x$.
    This inequality is true at some open interval containing zero except possibly at zero itself. Thus, by the squeeze theorem,
    $\displaystyle x\to 0,-x\to 0$.
    Thus, the limit of that is also zero.
    That means the function is continous.

    Similarly,
    $\displaystyle \lim_{x\to 0}x^2\cos (1/x)=0$
    Thus, it is continous.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Now we need to determine the derivative of $\displaystyle f,g$ at $\displaystyle x=0$.

    The derivative (if it exists) for $\displaystyle f$ is given by,
    $\displaystyle \lim_{\Delta x\to 0}\frac{(0+\Delta x)\sin (1/(0+\Delta x)) - f(0)}{\Delta x}$
    Which is equivalent to after simplification,
    $\displaystyle \lim_{\Delta x\to 0}\sin \frac{1}{\Delta x}$.
    This is a famous limits, which does not exist.

    However, if we used the second function after some manipulation, we would eventually reach after doing the same steps above,
    $\displaystyle \lim_{\Delta x\to 0}\Delta x\cos \frac{1}{\Delta x}=0$
    By using the Squeeze theorem trick employed before.

    Thus, $\displaystyle f$ is not differenciable at $\displaystyle x=0$. While $\displaystyle g$ is differenciable at $\displaystyle x=0$. That immediatly answers the (c) the first function's derivative is not defined at the point, thus it is not continous there.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    The only thing remains is to show that the derivative of $\displaystyle g$ is continous at $\displaystyle x=0$.

    If $\displaystyle x\not =0$ then the derivative of $\displaystyle g$ exists and is,
    $\displaystyle g'=(x^2)'\cos \frac{1}{x}+x^2\left( \cos \frac{1}{x} \right)'$
    $\displaystyle g'=2x\cos \frac{1}{x}+x^2\left(\frac{1}{x^2} \right) \sin \frac{1}{x}$
    $\displaystyle g'=2x\cos \frac{1}{x}+\sin \frac{1}{x}$.
    I do not think the limit $\displaystyle x\to 0$ exists. Because the first summand exists (by squeeze theorem) but the second one does not.

    Thus, the derivative is not continous.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by sbsite View Post
    thanks ?
    $\displaystyle
    f(x)=\left\{ {e^{-1/x^2} \ \ ,x \ne0 \atop 0 \ \ \ \ \ \ \ ,x=0} \right.
    $

    This is a well known function which has the curious property that all its
    derivatives at $\displaystyle x=0$ are $\displaystyle 0$. Showing this depends on showing that:

    $\displaystyle
    \lim_{x \to \infty} x^n e^{-x^2}=0
    $

    for all $\displaystyle n>0$, which I will leave as an excercise to the reader,
    it can be shown by repeated application of L'Hopital's rule. Alternativly you may
    be in the position of knowing that $\displaystyle e^{-x}$ decreases faster than any power of
    $\displaystyle x$, which is also sufficient to justify the above limit.

    $\displaystyle
    f'(x)=\frac{2}{x^3}e^{-1/x^2},\ x \ne 0
    $

    and:

    $\displaystyle
    f'(0)=\lim_{x \to 0} \frac{2}{x^3}e^{-1/x^2}
    $

    Put $\displaystyle u=1/x$, then:

    $\displaystyle
    f'(0)=\lim_{u \to \infty} 2\, u^3\,e^{-u^2}=0
    $

    The second derivative is similar, but a bit more fiddly.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by sbsite View Post
    thanks ?
    Problem 20, show:

    $\displaystyle
    x-(x^3/6) < \sin(x) < x,\ \ \ 0<x<\pi
    $

    This is a result of taking the Taylor expansion of $\displaystyle \sin(x)$ up to the term in $\displaystyle x^2$, and keeping the remainder term in:

    $\displaystyle
    \sin(x)=\sin(0)+\cos(0)x-\sin(0)x^2/2!-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)
    $


    $\displaystyle
    \sin(x)=x-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)
    $

    Now the required inequality follows by using the upper and lower bounds on $\displaystyle \cos(\zeta)$ for $\displaystyle \zeta \in [0,\pi]$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, sbsite!

    Here's #11 . . . but check my work, please.


    11) Find $\displaystyle \frac{dy}{dx}\bigg|_{x=0}$ for: .$\displaystyle y \;=\;\left[\ln(3x+2)\right]^{\sin^{-1}(2x+0.5)}$

    Take logs: .$\displaystyle \ln(y) \;=\;\ln\bigg\{\left[\ln(3x+2)\right]^{\sin^{-1}(2x+0.5)}\bigg\}\;=\;\sin^{-1}(2x+0.5)\cdot\ln\left[\ln(3x+3)\right]
    $

    Differentiate implicitly:

    . . $\displaystyle \frac{1}{y}\left(\frac{dy}{dx}\right)\:=\:\sin^{-1}(2x+0.5)\cdot\frac{1}{\ln(3x+2)}\cdot\frac{1}{3x +2}\cdot3$$\displaystyle + \frac{2}{\sqrt{1 - (2x+0.5)^2}}\cdot\ln\left[\ln(3x+2)\right] $

    We have: .$\displaystyle \frac{dy}{dx}\;=\;y\bigg\{\frac{3\sin^{-1}(2x+0.5)}{(3x+2)\ln(3x+2)} + \frac{2\ln\left[\ln(3x+2)\right]}{\sqrt{1 - (2x+0.5)^2}}\bigg\}
    $


    When $\displaystyle x = 0\!:\;y \:=\:\left(\ln 2\right)^{\sin^{-1}(0.5)} \:=\:\left(\ln 2\right)^{\frac{\pi}{6}} $

    Hence: .$\displaystyle \frac{dy}{dx}\:=\:(\ln 2)^{\frac{\pi}{6}}\left[\frac{3\cdot\sin^{-1}(0.5)}{2\cdot\ln2} + \frac{2\ln(\ln 2)}{\sqrt{1 - (0.5)^2}}\right] $

    . . . . . . . . $\displaystyle =\:(\ln 2)^{\frac{\pi}{6}}\left[\frac{3\cdot\frac{\pi}{6}}{2\ln 2} + \frac{2\ln(\ln 2)}{\sqrt{1-0.25}}\right]$

    . . . . . . . . $\displaystyle = \:(\ln2)^{\frac{\pi}{6}}\left[\frac{\pi}{4\ln2} + \frac{2\ln(\ln2)}{\sqrt{0.75}}\right] $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 3rd 2011, 05:35 AM
  2. binomial problems/normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Oct 19th 2010, 11:46 PM
  3. binomial problems as normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Oct 19th 2010, 11:41 PM
  4. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 25th 2010, 05:39 PM
  5. 2 Problems, need help, story problems!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 23rd 2007, 10:43 PM

Search Tags


/mathhelpforum @mathhelpforum