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Math Help - Evaluate the Limit

  1. #1
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    Evaluate the Limit

    Question : Evaluate

     <br />
\lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}<br />
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : Evaluate

     <br />
\lim_{x \to 0} \ (Cos x)^{\frac{1}{x^2}}<br />
    Since (\cos x)^{\frac{1}{x^2}} = e^{\frac{\ln \cos x}{x^2}}, consider \lim_{x \to 0} \frac{\ln \cos x}{x^2} (I suggest using l'Hopital's Rule).

    If you need more help, please show all your working and state specifically where you're stuck.
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  3. #3
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    as x\to0 we know that \frac{\ln(1-x)}x\to-1 and \frac{1-\cos x}{x^2}\to\frac12.

    now following up on mr fantastic work, we have that for x\ne0 is \frac{\ln (\cos x)}{x^{2}}=\frac{\ln \big(1-(1-\cos x)\big)}{1-\cos x}\cdot \frac{1-\cos x}{x^{2}}, and as x\to0 the whole limit is \frac1{\sqrt e}.
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  4. #4
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    Could u please explain me

    Quote Originally Posted by Krizalid View Post
    as x\to0 we know that \frac{\ln(1-x)}x\to-1 and \frac{1-\cos x}{x^2}\to\frac12.


    Could u please explain me how did u jump to the conclusion that \frac{\ln(1-x)}x\to-1 and \frac{1-\cos x}{x^2}\to\frac12.


    I am a novice so if u could provide me with the link where this can be derived i would appreciate it a lot
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  5. #5
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    Hint: Use l'hopital's on both limits.
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  6. #6
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    which 2 limits

    Quote Originally Posted by Defunkt View Post
    Hint: Use l'hopital's on both limits.

    could u tell me which two limits ....i can see only 1 limit...
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  7. #7
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    Quote Originally Posted by zorro View Post
    could u tell me which two limits ....i can see only 1 limit...
    For goodness sake! Go back and read post #4, your post. The one in which you asked about two limits. That is the post that Defunkt replied to.
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  8. #8
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    there is only on limit

    Quote Originally Posted by mr fantastic View Post
    For goodness sake! Go back and read post #4, your post. The one in which you asked about two limits. That is the post that Defunkt replied to.

    Sorry for annoying u so much but i cant understand

    #First.... how you got

    ,



    #Second..... how Krizalid concluded

    As we know that and
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  9. #9
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    mmm, that's serious.

    first one is a known property, it's just a=e^{\ln a}.

    and those limits, pretty basic, you can even prove them (if that pleases you) by applying L'H˘pital's rule.
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  10. #10
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    Thanks but what about the second one

    Quote Originally Posted by Krizalid View Post
    mmm, that's serious.

    first one is a known property, it's just a=e^{\ln a}.

    and those limits, pretty basic, you can even prove them (if that pleases you) by applying L'H˘pital's rule.

    Thanks for ur reply but what about the second one , could u please tel me how u derived that
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  11. #11
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    Quote Originally Posted by zorro View Post
    Thanks for ur reply but what about the second one , could u please tel me how u derived that
    There are several options for finding  \lim_{x \to 0} \frac{1 - \cos x}{x^2} but the simplest option is the best one for you to use: Apply l'Hopital's Rule twice.
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  12. #12
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    Is this correct

    Quote Originally Posted by mr fantastic View Post
    Since (\cos x)^{\frac{1}{x^2}} = e^{\frac{\ln \cos x}{x^2}}, consider \lim_{x \to 0} \frac{\ln \cos x}{x^2} (I suggest using l'Hopital's Rule).

    If you need more help, please show all your working and state specifically where you're stuck.

    \lim_{x \to 0} \frac{ln \ cosx}{x^2} = \lim_{x \to 0} \frac{f(x)}{g(x)}

    Using L'Hospitals Rule weget

    \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)}

    \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{- tanx}{2x}

    Dividing nu & de by x weget

    = \lim_{x \to 0} \frac{- \frac{tanx}{x}}{2}

    =  - \frac{ 1}{2}

    Is this correct or have i done something wrong.....And after this what should i do??
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  13. #13
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    Quote Originally Posted by zorro View Post
    \lim_{x \to 0} \frac{ln \ cosx}{x^2} = \lim_{x \to 0} \frac{f(x)}{g(x)}

    Using L'Hospitals Rule weget

    \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)}

    \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{- tanx}{2x}

    Dividing nu & de by x weget

    = \lim_{x \to 0} \frac{- \frac{tanx}{x}}{2}

    =  - \frac{ 1}{2}

    Is this correct or have i done something wrong.....And after this what should i do??
    Correct.

    So \lim_{x \to 0} (\cos x)^{\frac{1}{x^2}} = \lim_{x \to 0} e^{\frac{\ln \cos x}{x^2}} = e^{\lim_{x \to 0}{\frac{\ln \cos x}{x^2}}} = ....
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  14. #14
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    Thanks u very much

    Thank u Mr fantastic and every body else for ur guidance
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