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amoodarya

  • one year ago

The way to answer this limit

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  1. amoodarya
    • one year ago
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    \[\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\]

  2. anonymous
    • one year ago
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    try direct substitution

  3. anonymous
    • one year ago
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    i think every thing will be 0

  4. anonymous
    • one year ago
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    the answer =0

  5. amoodarya
    • one year ago
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    answer is +1

  6. amoodarya
    • one year ago
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    I can do that in very long terms ... but I looking for a good idea !

  7. anonymous
    • one year ago
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    ok but it is not undefined by direct substitution but 0

  8. freckles
    • one year ago
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    maybe we can use this... \[u=\sin(x)^{\cos(x)} \\ \frac{u'}{u}=\frac{\cos^2(x)}{\sin(x)}-\sin(x) \ln(\sin(x))\]

  9. ganeshie8
    • one year ago
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    then we simply need to find \(\lim\limits_{x\to 0}~~u' \) is it http://www.wolframalpha.com/input/?i=lim+%28x%5Cto+0%29+%28%28sin%28x%29%29%5E%28cos%28x%29%29%29%27

  10. freckles
    • one year ago
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    i was thinking about trying to somehow use l'hosptial backwards if you know what I mean

  11. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0} \frac{u'}{1}=\lim_{x \rightarrow 0} \frac{u}{x}\]

  12. freckles
    • one year ago
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    we need to show this \[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}=1\]

  13. freckles
    • one year ago
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    oops u does goes to 0

  14. amoodarya
    • one year ago
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    \[\lim sinx^{\cos x} \rightarrow 0 (x \to 0)\]

  15. thomas5267
    • one year ago
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    \[ \begin{align*} L&=\lim_{x\to 0} \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right)\\&=\lim_{x\to 0} \cos^2(x)\sin(x)^{\cos(x)-1} - \lim_{x\to 0} \sin(x)^{\cos(x)+1}\log(\sin(x))\\ \end{align*} \] By Mathematica, both limits exists. No idea how to evaluate though.

  16. thomas5267
    • one year ago
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    The first one is 1 and the second one is 0.

  17. thomas5267
    • one year ago
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    In fact, \[ \lim_{x\to 0}\sin(x)^{\cos(x)-1}=1 \]

  18. freckles
    • one year ago
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    \[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)}\] hmmm near x we have sin(x) is approximately x I don't know if we can do this...\[\lim_{x \rightarrow 0}\frac{1}{x} \sin(x)^{\cos(x)} =\lim_{x \rightarrow 0}\frac{1}{x} x ^{\cos(x)} =\lim_{x \rightarrow 0}x^{\cos(x)-1}\]

  19. freckles
    • one year ago
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    near x=0*

  20. thomas5267
    • one year ago
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    Squeeze theorem using x and -x?

  21. freckles
    • one year ago
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    \[v=x^{\cos(x)-1} \\ \ln(v)=(\cos(x)-1) \ln(x) \\ \lim_{x \rightarrow 0} (\cos(x)-1)\ln(x)=\lim_{x \rightarrow 0}\frac{\ln(x)}{\frac{1}{\cos(x)-1}} \\ =\lim_{x \rightarrow 0}\frac{\frac{1}{x}}{\frac{-\sin(x)}{(\cos(x)-1)^2}} =\lim_{x \rightarrow 0}\frac{-1(\cos(x)-1)^2}{x \sin(x)} \\ = \lim_{x \rightarrow 0} \frac{-1 (\frac{\cos(x)-1}{x})^2}{\frac{x}{x} \frac{\sin(x)}{x}}\]

  22. freckles
    • one year ago
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    \[=\frac{-1(0)^2}{1(1)}=0\]

  23. freckles
    • one year ago
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    \[e^{\ln(v)}->e^{0}=1\]

  24. freckles
    • one year ago
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    so v->1

  25. amoodarya
    • one year ago
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    thank U all . I get it

  26. thomas5267
    • one year ago
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    The problem is that \[ \lim_{x\to 0}\sin(x)^{\cos(x)-1} \quad\text{ does not exist on }\mathbb{R}\\ \lim_{x\to 0^+}\sin(x)^{\cos(x)-1}=1 \]

  27. thomas5267
    • one year ago
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    I don't think the left hand limit exists on real number.

  28. freckles
    • one year ago
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    thomas I see what you mean we need x>0 for ln(x) to exist also I think this approach might be better than my earlier if it is allowed \[\sin(x)\approx x \text{ for } x \approx 0 \\ \cos(x) \approx 1 \text{ for } x \approx 0 \\ \sin(x)^{\cos(x)}(\frac{\cos^2(x)}{\sin(x)}-\sin(x)\ln(\sin(x))) \approx x^{1}(\frac{1^2}{x}-x \ln(x)) \\ \lim_{x \rightarrow 0^+} x(\frac{1}{x}-x \ln(x)) \\ \\ =\lim_{x \rightarrow 0^+}(1-x^2 \ln(x))\]

  29. freckles
    • one year ago
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    like we can use l'hospital on that second term to show it goes to 0

  30. freckles
    • one year ago
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    then we have 1-0 which is 1

  31. thomas5267
    • one year ago
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    I mean \[ \sin(x)^{\cos(x)}\left(\frac{\cos^2(x)}{\sin(x)} - \sin(x)\log(\sin(x))\right) \; \text{does not exist for } x\leq 0 \]

  32. freckles
    • one year ago
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    right

  33. thomas5267
    • one year ago
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    So there is no left hand limit and no limit. At least not on real numbers.

  34. freckles
    • one year ago
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    but we can we find the right hand limit

  35. freckles
    • one year ago
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    I changed those little thingys above to + signs to denote I was looking at the right hand limit

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