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Math Help - A simple question based on limits

  1. #1
    iamtheone
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    A simple question based on limits

    i know the theorem that,
    when lim x tends to 0, then (sinx)/x=1.
    then my question is

    when lim x tends to 0, then what is sinx ?

    is the answer 0 or x?



    (i just know that we can multiply and divide by x. then (sinx)/x is 1 by theorem, then what will happen to the x which is multiplied in the numerator ? Limit will be also applied to that and the answer of the question would be 0. isn't it ? is it right or wrong ? )

    please tell me !!!!!!
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by iamtheone View Post
    i know the theorem that,
    when lim x tends to 0, then (sinx)/x=1.
    then my question is

    when lim x tends to 0, then what is sinx ?

    is the answer 0 or x?



    (i just know that we can multiply and divide by x. then (sinx)/x is 1 by theorem, then what will happen to the x which is multiplied in the numerator ? Limit will be also applied to that and the answer of the question would be 0. isn't it ? is it right or wrong ? )

    please tell me !!!!!!
    Let's look at the region of x \in [0, \frac{\pi}{2}]

    sin(x) \le x \le tan(x)

    \frac{sin(x)}{sin(x)} \le \frac{x}{sin(x)} \le \frac{tan(x)}{sin(x)}


    1 \le \frac{x}{sin(x)} \le \frac{sin(x)}{sin(x)cos(x)}


    \frac{x}{sin(x)} \le \frac{1}{cos(x)}

    lim_{x \rightarrow 0} \frac{1}{cos(x)} = 1

    Using the squeeze theorem:

    lim_{x \rightarrow 0} \frac{x}{sin(x)} = 1
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