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Math Help - help with some problems?

  1. #1
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    help with some problems?

    thanks ?
    Attached Thumbnails Attached Thumbnails help with some problems?-untitled-2.gif  
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  2. #2
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    We need to find the limits,
    \lim_{x\to 0} x\sin \frac{1}{x}
    We can write,
    -1\leq \sin \frac{1}{x} \leq 1
    Multiply by "x" (positive or negative equality is preseved),
    -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,
    x\to 0,-x\to 0.
    Thus, the limit of that is also zero.
    That means the function is continous.

    Similarly,
    \lim_{x\to 0}x^2\cos (1/x)=0
    Thus, it is continous.
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  3. #3
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    Now we need to determine the derivative of f,g at x=0.

    The derivative (if it exists) for f is given by,
    \lim_{\Delta x\to 0}\frac{(0+\Delta x)\sin (1/(0+\Delta x)) - f(0)}{\Delta x}
    Which is equivalent to after simplification,
    \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,
    \lim_{\Delta x\to 0}\Delta x\cos \frac{1}{\Delta x}=0
    By using the Squeeze theorem trick employed before.

    Thus, f is not differenciable at x=0. While g is differenciable at x=0. That immediatly answers the (c) the first function's derivative is not defined at the point, thus it is not continous there.
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  4. #4
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    The only thing remains is to show that the derivative of g is continous at x=0.

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

    Thus, the derivative is not continous.
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  5. #5
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    Quote Originally Posted by sbsite View Post
    thanks ?
    <br />
f(x)=\left\{ {e^{-1/x^2} \ \ ,x \ne0 \atop 0 \ \ \ \ \ \ \ ,x=0} \right.<br />

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

    <br />
\lim_{x \to \infty} x^n e^{-x^2}=0<br />

    for all 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 e^{-x} decreases faster than any power of
    x, which is also sufficient to justify the above limit.

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

    and:

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

    Put u=1/x, then:

    <br />
f'(0)=\lim_{u \to \infty} 2\, u^3\,e^{-u^2}=0<br />

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

    RonL
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by sbsite View Post
    thanks ?
    Problem 20, show:

    <br />
x-(x^3/6) < \sin(x) < x,\ \ \ 0<x<\pi<br />

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

    <br />
\sin(x)=\sin(0)+\cos(0)x-\sin(0)x^2/2!-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)<br />


    <br />
\sin(x)=x-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)<br />

    Now the required inequality follows by using the upper and lower bounds on \cos(\zeta) for \zeta \in [0,\pi].
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  7. #7
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    Hello, sbsite!

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


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

    Take logs: . \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]<br />

    Differentiate implicitly:

    . . \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  + \frac{2}{\sqrt{1 - (2x+0.5)^2}}\cdot\ln\left[\ln(3x+2)\right]

    We have: . \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\}<br />


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

    Hence: . \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]

    . . . . . . . . =\:(\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]

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

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