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anonymous

  • one year ago

Trigonometric limit help.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    I can't figure out how to fix the denominator in any way.

  3. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0} \frac{x^3 \sin(2x)}{\cos(x^4-\frac{\pi}{2}) \tan(x^3+x)}\] writing it here so I don't have to keep looking back and forth and now thinking...

  4. anonymous
    • one year ago
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    Yep no problems haha.

  5. anonymous
    • one year ago
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    Take a look at my work too. I figured out almost all of it.

  6. freckles
    • one year ago
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    what happens when you plug in 0 we should get 0/0 so we definitely have more work to do...

  7. anonymous
    • one year ago
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    It's just that pesky first limit.

  8. anonymous
    • one year ago
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    If I was allowed to use taylor polynomials this would be a joke but I am not.

  9. freckles
    • one year ago
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    yeah I was looking at your work and that one term we have a problem with since that first limit doesn't actually exist \[\lim_{x \rightarrow 0} \frac{x^3\cos(x^3+x)(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x} \\ \text{ plug \in 0 you get } \frac{1}{0} \\ \text{ so the limit does not exist } \\ \text{ so I don't think we can take the limit of each factor }\]

  10. freckles
    • one year ago
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    you get 2/0 sorry forgot to multiply the 2 on top to the 1 on top

  11. freckles
    • one year ago
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    still doesn't exist though

  12. anonymous
    • one year ago
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    According to wolfram, it's 4 though.

  13. freckles
    • one year ago
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    for the one I just said doesn't exist?

  14. anonymous
    • one year ago
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    WAIT.

  15. anonymous
    • one year ago
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    OMG> Sorry correction. The sin(2x) in the numerator should be a sin^2(2x) .

  16. freckles
    • one year ago
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    so the work on that one page is slightly off then because of this new factor sin(2x)

  17. anonymous
    • one year ago
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    Yep let me try again...

  18. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0} \frac{ x^3 \cos(x^3+x) (2x) \sin(2x)}{x^4(x^3+x)} \\ \lim_{x \rightarrow 0} \frac{\sin(2x)}{2x} (2) \frac{x^3 \cos(x^3+x)(2x)}{x^3(x^3+x) } \\ \lim_{x \rightarrow 0} 2 \frac{\cos(x^3+x)}{x^3+x}(2x) \\ \lim_{x \rightarrow 0 } 4 \frac{\cos(x^3+x)}{x^2+1}\] plug in 0 :)

  19. freckles
    • one year ago
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    and dido awesome algebra going on your work

  20. anonymous
    • one year ago
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    Omg thanks T_T . I can't believe a typo threw me off.

  21. anonymous
    • one year ago
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    Haha thanks!

  22. freckles
    • one year ago
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    yep throwing in that sin(2x) into your first factor there totally fixed it so the limit would exist

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