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Math Help - Textbook discrepancy

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Textbook discrepancy

    Today in one of my texts I saw this

    \text{As}x\to{0},\tan(x^3)\sim{x^2}

    I dont think this is right?

    Correct me if I am wrong but Three ways prove I am right unless I am missing somethign stupid

    \lim_{x\to{0}}\frac{\tan(x^3)}{x^2}=\lim_{x\to{0}}  \frac{\frac{\sin(x^3)}{\cos(x^3)}}{x^2}=\lim_{x\to  {0}}\frac{sin(x^3)}{\cos(x^3)\cdot{x^2}}

    Now I will apply that \sin(x^3)\sim{x^3}\text{As}x\to{0}

    To rewrite this as \lim_{x\to{0}}\frac{x^3}{\cos(x^3)\cdot{x^2}}=\lim  _{x\to{0}}\frac{x}{\cos(x^3)}=0\ne{1}

    You can also do this limit through power series and L'hopital's.

    Where am I making my mistake?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Today in one of my texts I saw this

    \text{As}x\to{0},\tan(x^3)\sim{x^2}

    I dont think this is right?

    Correct me if I am wrong but Three ways prove I am right unless I am missing somethign stupid

    \lim_{x\to{0}}\frac{\tan(x^3)}{x^2}=\lim_{x\to{0}}  \frac{\frac{\sin(x^3)}{\cos(x^3)}}{x^2}=\lim_{x\to  {0}}\frac{sin(x^3)}{\cos(x^3)\cdot{x^2}}

    Now I will apply that \sin(x^3)\sim{x^3}\text{As}x\to{0}

    To rewrite this as \lim_{x\to{0}}\frac{x^3}{\cos(x^3)\cdot{x^2}}=\lim  _{x\to{0}}\frac{x}{\cos(x^3)}=0\ne{1}

    You can also do this limit through power series and L'hopital's.

    Where am I making my mistake?
    tan(x^3) \approx x^3 for small x. Just find the Maclaurin series. No heavy lifting with the limits is required.

    -Dan
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by topsquark View Post
    tan(x^3) \approx x^3 for small x. Just find the Maclaurin series. No heavy lifting with the limits is required.

    -Dan
    Yeah that is what I said...and of course I did use Maclaurin ..so the book is wrong then for sure?
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    \sin(x^3)\sim{x^3}\text{As}x\to{0}
    Please elaborate.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by hatsoff View Post
    Please elaborate.
    This notation means that as x\to{0} then \sin(x^{n})\to{x^n}

    Really what it means is as x approaches 0 \sin(x^n) and x^{n} become equivalent
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    This notation means that as x\to{0} then \sin(x^{n})\to{x^n}

    Really what it means is as x approaches 0 \sin(x^n) and x^{n} become equivalent
    Okay, so, in other words:

    \lim_{x\to{0}}\frac{sinx^3}{x^3}=1

    But how do you determine that it really does equal 1? (I don't know how to work with trig limits.)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by hatsoff View Post
    Okay, so, in other words:

    \lim_{x\to{0}}\frac{sinx^3}{x^3}=1

    But how do you determine that it really does equal 1? (I don't know how to work with trig limits.)
    Two ways

    L'hopital's

    \lim_{x\to{0}}\frac{\sin(x^3)}{x^3}=\lim_{x\to{0}}  \frac{\cos(x^3)\cdot{3x^2}}{3x^2}=\lim_{x\to{0}}\c  os(x^3)=1

    Or inserting the Macluarin series you get

    \lim_{x\to{0}}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...}{x^3}=\lim_{x\to{0}}1-\frac{x^{6}}{3!}+\frac{x^{12}}{5!}-...=1
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    Two ways

    L'hopital's

    \lim_{x\to{0}}\frac{\sin(x^3)}{x^3}=\lim_{x\to{0}}  \frac{\cos(x^3)\cdot{3x^2}}{3x^2}=\lim_{x\to{0}}\c  os(x^3)=1

    Or inserting the Macluarin series you get

    \lim_{x\to{0}}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...}{x^3}=\lim_{x\to{0}}1-\frac{x^{6}}{3!}+\frac{x^{12}}{5!}-...=1
    Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.

    Thanks.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by hatsoff View Post
    Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.

    Thanks.
    I love Maclaurin series...I actually just created a long and hopefully enlightening tutorial on them today. I posted it like an hour ago but it might be gone for review
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by hatsoff View Post
    Hmph. So that's twice in one night that Maclaurin series have reared their heads. I will have to make myself more familiar with them; presently I find them to be quite unwieldy, for lack of practice/use.
    If you don't know MacLaurin series, you can use the definition of the derivative of the sine taken at 0 :

    <br />
\lim_{x\to0}\frac{\sin x^3}{x^3}=\lim_{X\to0} \frac{\sin X-\sin 0}{X-0} =\sin'(0)=\cos 0=1
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