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Math Help - How to evaluate this limit?

  1. #1
    Member zzizi's Avatar
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    How to evaluate this limit?

    I need to evaluate the lim x->0 tan^2(x)/x

    I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

    It could be re-written as:
    (sinx/cosx)(sinx/cosx)/x

    The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: How to evaluate this limit?

    You could use L'H˘pital's rule to write the limit as:

    2\lim_{x\to0}\tan(x)\sec^2(x)=0
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  3. #3
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    Re: How to evaluate this limit?

    Hi zzizi,

    lim x->0 sin(x)/x =1,
    lim x->0 tan(x)/x =1,
    lim x->0 tan(x)/x *tan(x) = 1*0 = 0
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  4. #4
    Member zzizi's Avatar
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    Re: How to evaluate this limit?

    Thank you both very much!
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  5. #5
    Forum Admin topsquark's Avatar
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    Re: How to evaluate this limit?

    Quote Originally Posted by zzizi View Post
    It could be re-written as:
    (sinx/cosx)(sinx/cosx)/x

    The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?
    Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

    -Dan
    Thanks from MarkFL
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  6. #6
    MHF Contributor
    Prove It's Avatar
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    Re: How to evaluate this limit?

    Quote Originally Posted by zzizi View Post
    I need to evaluate the lim x->0 tan^2(x)/x

    I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

    It could be re-written as:
    (sinx/cosx)(sinx/cosx)/x

    The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?
    \displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}
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  7. #7
    Member zzizi's Avatar
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    Re: How to evaluate this limit?

    Quote Originally Posted by topsquark View Post
    Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

    -Dan
    Yes, perhaps I was a bit hasty there.
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  8. #8
    Member zzizi's Avatar
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    Re: How to evaluate this limit?

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}
    Wonderful! thank you sir!
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