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Basic math doubt

  1. Jun 11, 2012 #1
    can you explain why this happen? i can't understand

    limite.png

    thks in advance
     
  2. jcsd
  3. Jun 11, 2012 #2
    They're kinda the same thing...

    [itex]\frac{1}{0} = \infty[/itex]
    [itex]\sin^2(0) = 0[/itex]

    so

    [itex]\frac{\sin^2(0)}{0} = \underbrace{\frac{0}{0}}_\text{ind.} = \frac{1}{0}\cdot\frac{0}{1} = \underbrace{\infty\cdot0}_\text{ind.} = \frac{1}{0}\cdot\frac{\sin^2(0)}{1}[/itex]

    Use l'Hôpital's rule to find the value for this limit.

    If [itex]\lim\limits_{x\to\alpha}\frac{f(x)}{g(x)} = \frac{0}{0} \mathrm{or} \pm\frac{\infty}{\infty}[/itex], then [itex]\lim\limits_{x\to\alpha}\frac{f(x)}{g(x)} = \lim\limits_{x\to\alpha}\frac{f'(x)}{g'(x)}[/itex]

    so, define functions...

    [itex]f: x \ {\mapsto}\ \sin\left(\frac{1}{2} \, x\right)^{2}[/itex]

    [itex]g: x \ {\mapsto}\ x^{2}[/itex]

    Differentiate until you aren't going to get an indeterminate value (that is, make sure that the denominator does not equal zero).

    [itex]\frac{\mathrm{d}g}{\mathrm{d}x} = 2 \, x[/itex]

    [itex]\frac{\mathrm{d}^{2}g}{\mathrm{d}x^{2}} = 2[/itex]

    So now the denominator will never equal zero. Differentiate [itex]f[/itex] the same number of times.

    [itex]\frac{\mathrm{d}f}{\mathrm{d}x} = \sin\left(\frac{1}{2} \, x\right) \cos\left(\frac{1}{2}
    \, x\right)[/itex]

    [itex]\frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}} = -\frac{1}{2} \, \sin\left(\frac{1}{2} \, x\right)^{2} +
    \frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2}[/itex]

    Now divide the functions

    [itex] \frac{f''(x)}{g''(x)} = -\frac{1}{4} \, \sin\left(\frac{1}{2} \, x\right)^{2} +
    \frac{1}{4} \, \cos\left(\frac{1}{2} \, x\right)^{2} [/itex]

    simplifies to

    [itex] \frac{f''(x)}{g''(x)} = \frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -
    \frac{1}{4} [/itex]

    now take the limit as [itex]x[/itex] tends to zero

    [itex] \lim\limits_{x\to0}\frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -
    \frac{1}{4} = \frac{1}{2} \, \cos\left(0\right)^{2} -
    \frac{1}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} [/itex]

    And, to bring it all together:

    [itex]\lim\limits_{x\to0}\dfrac{\sin^{2}\left(\frac{x}{2}\right)}{x^{2}} = \lim\limits_{x\to0}\frac{1}{2} \, \cos\left(\frac{1}{2} \, x\right)^{2} -
    \frac{1}{4} = \frac{1}{4}[/itex]

    I hope that helps!
     
    Last edited: Jun 11, 2012
  4. Jun 12, 2012 #3
    I still don't understand why people want to divide by zero. It's completely nonsense...
     
  5. Jun 12, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Hey, Ive never understood why people want to eat Haggis!

    People continue to act irrationally!
     
  6. Jun 12, 2012 #5
    It's sorta completely nearly nonsense... If the limit of whatever/x as x->0 is the same from both sides, then I call that close enough as long as it's not part of some weird "1=0" abuse of math. It is possible that I just haven't taken enough classes to know that I'm wrong, though.
     
  7. Jun 12, 2012 #6
    Digital signal processing makes uses of poles and the [itex]\mathrm{sinc}(x)[/itex] function.

    The normalized [itex]\mathrm{sinc}(x)[/itex] function:

    [itex]
    \mathrm{sinc}(x) = \left\{
    \begin{array}{cc}
    1 & \text{if} \quad x = 0\\
    \frac{\sin(\pi \, x)}{\pi \, x} & \text{if} \quad x \neq 0
    \end{array}
    \right.
    [/itex]

    Has uses in DSP as its Fourier Transform is the rectangle function, which is the ideal low-pass filter.

    [itex]
    \mathrm{rect}(\xi) = \mathcal{F}_t\left[\frac{\sin(\pi \, x)}{\pi \, x}\right]\left(\xi\right)
    [/itex]

    Poles in filters determine the Q-factor, or bandwidth of the filter. I think, anyway. Or what frequencies a filter affects. I don't remember, I read some college DSP notes when high school trig was getting boring, so it went over my head a little.

    [itex]
    \dfrac{\omega^{3} - 17 \, \omega^{2} + 94 \, \omega - 168}{\omega^{3} - 14 \, \omega^{2} + 51 \, \omega
    - 54} = \dfrac{{\left(\omega - 7\right)} {\left(\omega - 6\right)} {\left(\omega -
    4\right)}}{{\left(\omega - 9\right)} {\left(\omega - 3\right)} {\left(\omega -
    2\right)}}
    [/itex]

    The above has zeros at [itex]\omega = \left[7, 6, 4\right][/itex] and poles at [itex]\omega = \left[9, 3, 2\right][/itex].

    Division by zero has its uses. I'm not sure outside of DSP, though.
     
  8. Jun 12, 2012 #7
    You're not actually dividing by zero. You come close to dividing by zero, never do, and the ratios of the two functions that are heading to zero, that ratio may actually head towards a well-defined number.

    Sorry to burst your "ooh, there's a rip in the space-time continuum" bubble.
     
  9. Jun 12, 2012 #8
    This is true. But we are talking about limits, so I figured that that was implied.
     
  10. Jun 12, 2012 #9
    Again, there's no dividing by 0, just taking a limit. So with the greatest possible respect, shaddap. There's a lot of stuff taking limits is useful for, and if plugging in happens to result in 0/0, it's just one you can't solve by just plugging in.
     
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