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anonymous

  • one year ago

limit x tends to 0 for f(x)= (cos(sinx)-cosx)/x^2

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  1. anonymous
    • one year ago
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    I've tried several things, transformation formulae, series expansions etc etc. Very stuck.

  2. Jhannybean
    • one year ago
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    \[\large \lim_{x\rightarrow 0 } \frac{\cos(\sin(x))-\cos(x)}{x^2}\]

  3. anonymous
    • one year ago
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    On mobile, so thanks for the formatting.

  4. Astrophysics
    • one year ago
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    Split it into two limits, one should require squeeze theorem I believe

  5. anonymous
    • one year ago
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    Squeeze theorem doesn't make sense if you have a x^2 term I thought?

  6. Jhannybean
    • one year ago
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    No, it does.

  7. ganeshie8
    • one year ago
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    we cannot split the limit here because the individual limits don't exist

  8. ganeshie8
    • one year ago
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    have u tried taylor ? looks the numerator is \(\mathcal{O}(x^3)\)

  9. Astrophysics
    • one year ago
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    I think you maybe right no we can just use L'hopital's rule

  10. Astrophysics
    • one year ago
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    0/0

  11. imqwerty
    • one year ago
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    what i did - 1) simplified cosA+cosB 2)took the constants out ...... only constant which ws there ws -2 :P 3)applied L hospitality nd then i got (-2)x0 =0 :)

  12. Astrophysics
    • one year ago
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    looks good @qwerty!

  13. imqwerty
    • one year ago
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    :) thanks @Astrophysics

  14. Jhannybean
    • one year ago
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    cos(A) \(\color{red}{\text{+}}\) cos(B)?

  15. Empty
    • one year ago
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    Yeah I did L'H twice and got 0 as well

  16. Astrophysics
    • one year ago
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    Yup did same thing haha

  17. anonymous
    • one year ago
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    Oh well, we're required to solve without L'Hospital. So anyone got any clues on that.

  18. Empty
    • one year ago
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    Any other restrictions we should know about...?

  19. Empty
    • one year ago
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    If you can use series expansions then saying you can't use L'H is practically meaningless imo

  20. anonymous
    • one year ago
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    And I don't know what or how Taylor Series are... No other restrictions.

  21. ganeshie8
    • one year ago
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    series expansion works nicely

  22. ganeshie8
    • one year ago
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    just show that the degree of numerator is at least 3 and you're done

  23. Astrophysics
    • one year ago
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    Niceee

  24. anonymous
    • one year ago
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    Okay? @ganeshie8 and that's proof cause?

  25. ganeshie8
    • one year ago
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    so are you allowed to use below ? \[\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots\] \[\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots\]

  26. anonymous
    • one year ago
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    yes.

  27. ganeshie8
    • one year ago
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    then use them

  28. anonymous
    • one year ago
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    I've gone the next few steps... but I don't understand how that gives me a zero.!?!?

  29. ganeshie8
    • one year ago
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    \[\large \cos(\sin x) = 1-\dfrac{(\sin x)^2}{2}+\cdots = 1 -\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\cdots \] fine with above ?

  30. anonymous
    • one year ago
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    yes. got that step. Im afraid I need to be led by the hand still though.

  31. ganeshie8
    • one year ago
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    \[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{1-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{-\dfrac{x^2}{2}+\mathcal{O}(x^4)}\right) \\~\\ &= \color{red}{-\dfrac{\mathcal{O}(x^4)}{2}+\mathcal{O}(x^4)}-\left(\color{blue}{\mathcal{O}(x^4)}\right) \\~\\ &=\mathcal{O}(x^4) \end{align}}\]

  32. ganeshie8
    • one year ago
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    that shows that the degree of each term in the numerator is at least \(4\)

  33. anonymous
    • one year ago
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    I'm so sorry, but whats that o like symbol? starting out with calc

  34. ganeshie8
    • one year ago
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    If it helps, you may replace \(\mathcal{O}(x^4)\) by \(x^4(stuff)\)

  35. anonymous
    • one year ago
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    got it. thanks

  36. anonymous
    • one year ago
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    this actually pretty clever.

  37. ganeshie8
    • one year ago
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    \[\large{\begin{align} &\color{red}{ \cos(\sin x)}-\color{blue}{\cos x}\\~\\ &= \color{red}{1-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{1-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{(x - \frac{x^3}{3!}+\cdots)^2}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^2+x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{-\dfrac{x^2}{2}+x^4(stuff)}\right) \\~\\ &= \color{red}{-\dfrac{x^4(stuff)}{2}+x^4(stuff)}-\left(\color{blue}{x^4(stuff)}\right) \\~\\ &=x^4(stuff) \end{align}}\]

  38. anonymous
    • one year ago
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    oh no need for that. I worked it out too. thanks a lot lot lot

  39. Astrophysics
    • one year ago
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    |dw:1440488705192:dw|

  40. anonymous
    • one year ago
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    I'm like that after every single calc question. so many...So many. I'm probably going to keep asking today

  41. ganeshie8
    • one year ago
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    |dw:1440489968673:dw|

  42. Astrophysics
    • one year ago
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    Cute xD

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