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amyna

  • one year ago

find the limit

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  1. amyna
    • one year ago
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    |dw:1444585400774:dw|

  2. amyna
    • one year ago
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    how do i do the l'hopitals rule for this problem?

  3. freckles
    • one year ago
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    I guess you are asking how to differentiate sin(pi*x) since I'm certain you know how to differentiate x

  4. amyna
    • one year ago
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    so far i got: 1+cos pi x / 1-cos pi x ?

  5. freckles
    • one year ago
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    \[\frac{d}{dx}(\sin(\pi x)) =(\pi x)' \cos(\pi x) \text{ by chain rule }\]

  6. freckles
    • one year ago
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    well (pi*x)'=pi not 1

  7. freckles
    • one year ago
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    anyways you should be able to use direct sub after the first round of l'hospital once you differentiate sin(pi*x) correctly

  8. amyna
    • one year ago
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    so pi x + cos pi x / pi x - cos pi x ?

  9. freckles
    • one year ago
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    derivative of just x is 1... why do you put pi x? you still aren't writting the derivative of sin(pi*x) is pi*cos(pi*x)

  10. freckles
    • one year ago
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    \[\frac{d}{dx}(x)=1 \\ \frac{d}{dx} \sin( \pi x)=(\pi x)' \cos(\pi x)= \pi \cos(\pi x)\]

  11. amyna
    • one year ago
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    oh okay so the answer is 1+ pi / 1- pi

  12. freckles
    • one year ago
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    yes that is right \[\lim_{x \rightarrow 0} \frac{x+ \sin( \pi x)}{x- \sin (\pi x)} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x + \sin(\pi x))}{\frac{d}{dx}(x- \sin(\pi x))} \\ \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x)+\frac{d}{dx}\sin(\pi x)}{\frac{d}{dx}(x)-\frac{d}{dx}\sin (\pi x)}\]

  13. freckles
    • one year ago
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    well almost it should be (1+pi)/(1-pi)

  14. amistre64
    • one year ago
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    1+2sin(px)/x ------------ x-sin(px) | x + sin(px) (x -sin(px)) ------------ 2sin(px) (2sin(px) -2sin^2(px)/x) --------------------- 2sin^2(px)/x sin(px) = 0 when x=0, so i spose we could view this as 1+2sin(px)/x as x to 0, maybe?

  15. amistre64
    • one year ago
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    i think my thought has an error tho

  16. freckles
    • one year ago
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    are you doing yours with l'hospital ? I was trying to follow all those broken line thingys...

  17. amistre64
    • one year ago
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    i was trying to do long division :) to see what a series might be represented as

  18. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0} \frac{x + \sin( \pi x)}{x- \sin( \pi x)} \\ \text{ divide top and bottom by } \pi x \\ \lim_{x \rightarrow 0} \frac{\frac{x}{\pi x}+\frac{\sin(\pi x)}{\pi x}}{\frac{x}{\pi x}-\frac{\sin(\pi x)}{\pi x}} \\ =\frac{\frac{1}{\pi}+1}{\frac{1}{\pi}-1} \\ \text{ now multiply top and bottom by } \pi \\ =\frac{1+\pi}{1-\pi}\]

  19. amistre64
    • one year ago
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    squashing my creativity are you .... ;)

  20. freckles
    • one year ago
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    lol no let's try the long division thingy

  21. freckles
    • one year ago
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    or let me look at yours since I know what I'm looking at :p

  22. amyna
    • one year ago
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    lol Thanks for you help guys! :)

  23. amistre64
    • one year ago
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    maybe fliiping the order of the terms might help .. one way is x to infinity, the other is x to zero ... ive noticed this but i cant see why.

  24. freckles
    • one year ago
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    your division seems great

  25. freckles
    • one year ago
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    so you get this : \[\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x)})\]

  26. amistre64
    • one year ago
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    i was thinking we could factor our a sin(px) to zero out all but the first 2 terms .. but that seems to be where my idea goes awry.

  27. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0} (1+2 \frac{\sin(\pi x)}{x}+2 \cdot \frac{\sin^2( \pi x)}{x(x- \sin( \pi x))})\]

  28. amistre64
    • one year ago
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    i think i broke the wolf :/

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  29. freckles
    • one year ago
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    \[1+2 \pi +2 \cdot \lim_{x \rightarrow 0} \frac{\sin^2(\pi x)}{x^2- x \sin(\pi x)} \\ \\ \text{ for the last quotient there divide top and bottom by } x^2 \\ 1+2 \pi +2 \cdot \frac{\pi^2}{1-\pi} \] I still go back to my old ways using that one famous limit thingy

  30. freckles
    • one year ago
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    I think I did something wrong though

  31. freckles
    • one year ago
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    oh no it was the same as previous answer

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