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najwaischua

  • 2 years ago

lim x->o (1-cosx)/x sqrt

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  1. myininaya
    • 2 years ago
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    This seems to be a bit weird sqrt of what?

  2. dpaInc
    • 2 years ago
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    it's the new way...

  3. FoolForMath
    • 2 years ago
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    lol

  4. Kreshnik
    • 2 years ago
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    LOL @myininaya I thought It was \[\LARGE \lim_{x\to 0}{1-\cos x \over \sqrt x}\]

  5. najwaischua
    • 2 years ago
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    the lower is x^2

  6. Kreshnik
    • 2 years ago
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    \[\LARGE \lim_{x\to 0}{1- \cos x\over x^2}\] ?

  7. najwaischua
    • 2 years ago
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    yes yes

  8. myininaya
    • 2 years ago
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    do you l'hospital?

  9. myininaya
    • 2 years ago
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    i mean know?

  10. najwaischua
    • 2 years ago
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    yes. but the question say use limit rule. don't use l'opitall's rule

  11. najwaischua
    • 2 years ago
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    i've tried to use factorization. but can't get it

  12. myininaya
    • 2 years ago
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    ok \[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x^2} \cdot \frac{1+\cos(x)}{1+\cos(x)}\] \[\lim_{x \rightarrow 0}\frac{1-\cos^2(x)}{x^2(1+\cos(x))}=\lim_{x \rightarrow 0}\frac{\sin^2(x)}{x^2(1+\cos(x))}\] need more help?

  13. myininaya
    • 2 years ago
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    \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x} \cdot \lim_{x \rightarrow 0}\frac{\sin(x)}{x} \cdot \lim_{x \rightarrow 0}\frac{1}{1+\cos(x)}\]

  14. najwaischua
    • 2 years ago
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    the answer should be?

  15. myininaya
    • 2 years ago
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    so you got this right ?

  16. najwaischua
    • 2 years ago
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    actually no. what technique is this?

  17. myininaya
    • 2 years ago
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    you wanted to use algebra and limit laws...

  18. myininaya
    • 2 years ago
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    do you have any questions with the steps i performed?

  19. niki
    • 2 years ago
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    is the ans 1/2?

  20. Kreshnik
    • 2 years ago
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    you're supposed to know this rule: \[ \lim_{x\to0}{\sin x\over x}= ?\] what should be instead of "?"

  21. najwaischua
    • 2 years ago
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    1

  22. Kreshnik
    • 2 years ago
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    so there's nothing to get confused of :) .. @myininaya solved it, you just had to substitute :)

  23. najwaischua
    • 2 years ago
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    thank you

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