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Math Help - help me in this kind of questions

  1. #1
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    help me in this kind of questions

    please solve this question and also show the steps of solution

    1) Lim x->0 x^(2sinx)

    ie lim x tends to 0, x to the power 2sinx

    2) is L-hospitals rule valid for 0 to the power 0 condition
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by ayush View Post
    please solve this question and also show the steps of solution

    1) Lim x->0 x^(2sinx)

    ie lim x tends to 0, x to the power 2sinx

    2) is L-hospitals rule valid for 0 to the power 0 condition
    Set y = x^{sin2x} and take the log of both sides:

    ln y = sin(2x) ln(x)
    ln y = \frac{ln(x)}{csc(2x)}

    The limit of the right hand side as x goes to zero can be found using L'Hopital's Rule. (verify conditions)
    So the limit as x tends to 0 of \frac{\frac{1}{x}}{-2\frac{cos(2x)}{sin^{2}(2x)}}
    \frac{sin^{2}(2x)}{-2xcos(2x)}

    Note that the limit of \frac{sin(2x)}{2x} is 1, so you're left with
    \frac{sin(2x)}{-cos(2x)}, whose limit is 0.

    But remember that we just measured the limit of ln y equal to 0, which means the limit of y, your original expression, is 1. Verify numerically and graphically.


    Good luck!!
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  3. #3
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    please help me in this question also

    lim
    x->infinity ((1+1/x^2)/x)

    please tell the steps also
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  4. #4
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by ayush View Post
    lim
    x->infinity ((1+1/x^2)/x)

    please tell the steps also
    You need to start a new thread. Please spend some time reading the forum rules.

    Start by simplifying the expression

    \frac{\frac{x^2+1}{x^2}}{x} = \frac{x^2+1}{x^3}

    as x blows up the ratio goes to zero because x^3 grows faster.
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  5. #5
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    Quote Originally Posted by ayush View Post
    lim
    x->infinity ((1+1/x^2)/x)

    please tell the steps also
    Just go ahead and do the multiplication: (1+ \frac{1}{x})\frac{1}{x}= \frac{1}{x}+ \frac{1}{x^2}. What happens to that as x goes to infinity?

    You could also do this "piece by piece". What is the limit of 1+ \frac{1}{x} as x goes to infinity? Now, what is the limit of that, divided by x, as x goes to infinity?
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