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karatechopper

  • one year ago

Find the limit. (Special Trig functions) As the lim (sinx/2x^2-x) approaches 0.

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  1. karatechopper
    • one year ago
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    I broke the limit down and did this: sinx sinx ---- - ----- x 2x-1 I used the special trig function to find out that sinx/x=1. So now I am left with 1 - sinx/2x-1 And I'm stuck!

  2. karatechopper
    • one year ago
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    to make more sense- I am stuck on 1- (sinx/2x-1)

  3. IrishBoy123
    • one year ago
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    \[lim_{x\rightarrow 0} \frac{sin \ x}{2x^2 - x} = lim_{x\rightarrow 0} \frac{\frac{sin \ x}{x}}{2x - 1} \] plus \[lim_{x\rightarrow 0} \frac{sin}{ x} = 1\]

  4. ChillOut
    • one year ago
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    Still having trouble?

  5. karatechopper
    • one year ago
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    yeah I still don't understand..

  6. IrishBoy123
    • one year ago
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    \(\large \frac{1}{2(0) - 1} = ???\)

  7. ChillOut
    • one year ago
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    You need to notice that 2x²-x = x(2x-1). That's what @IrishBoy123 did.

  8. ChillOut
    • one year ago
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    And that's the same as \(\frac{ \sin x }{ x }*\frac{ 1 }{ 2x-1 }\)

  9. IrishBoy123
    • one year ago
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    the crucial thing here is that \[lim_{x\rightarrow 0} \frac{sin \ x}{x} = 1\] so you set it up to have a \( \frac{sin \ x}{x} \) term but that \( \frac{sin \ x}{x} \) term is seminal

  10. IrishBoy123
    • one year ago
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    this the important bit @karatechopper ! https://gyazo.com/ad5c26d3cf8e2b41c20a8c7935b34a83 splitting it out so you have a \(\frac{sinx}{x}\) term

  11. anonymous
    • one year ago
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    @ChillOut hi

  12. karatechopper
    • one year ago
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    understood! thanks :)

  13. karatechopper
    • one year ago
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    BUT. How can you just take that x to the numerator when its being multiplied in the denominator?

  14. ChillOut
    • one year ago
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    When you have something like \(\LARGE\frac{\frac{a}{b}}{c}\) you can rewrite that as \(\LARGE\frac{a}{b*c}\).

  15. ChillOut
    • one year ago
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    @karatechopper

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