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Math Help - [SOLVED] Limit confusion: different methods, different solutions :S

  1. #1
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    [SOLVED] Limit confusion: different methods, different solutions :S

    Hey there

    note: x approaches 0 on the limit, didn't know how to format the code
    I was solving this problem on my textbook:
    lim((e^(x^2)-cos(x))/x^2)=?
    On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

    On the other hand, I tried this solution:

    From the limit laws we get:
    lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)
    we know that limcos(x) as x approaches 0 is 1, so we end up:
    lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

    Thanks
    Last edited by FreeedomOverdose; January 4th 2010 at 04:58 AM.
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  2. #2
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    Quote Originally Posted by FreeedomOverdose View Post
    Hey there

    note: The x approaches 0 on the limit, didn't know how to format the code
    I was solving this problem on my textbook:
    lim((e^(x^2)-cos(x))/x^2)=?
    On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

    On the other hand, I tried this solution:

    From the limit laws we get:
    lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)
    we know that limcos(x) as x approaches 0 is 1, so we end up:
    lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

    Thanks
    Your statment "we end up: lim((e^(x^2)- 1)/x^2)" is wrong. You have already taken the limit in one part of the formula but not the others- you can't do that.
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  3. #3
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    Hey

    Thanks for your quick reply. Actually, I mean this:

    lim(e^(x^2)-cos(x))=lim(e^(x^2))-lim(cos(x))= lim(e^(x^2))-1
    now, we replace 1 with lim(1) (as x approaches 0)
    so we have:
    lim(e^(x^2))-lim(1) => (sum of limits=limit of sum)lim((e^(x^2))-1)

    :S
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  4. #4
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    Quote Originally Posted by FreeedomOverdose View Post
    Hey there

    note: The x approaches 0 on the limit, didn't know how to format the code
    I was solving this problem on my textbook:
    lim((e^(x^2)-cos(x))/x^2)=?
    On my first method, I replaced cos(x) with 1-sin^2(x/2) and after some modifications, the result turned out to be 3/2 (which was also the result on the textbook).

    On the other hand, I tried this solution:

    From the limit laws we get:
    lim((e^(x^2)-cos(x))/x^2)=lim(e^(x^2))-lim(cos(x))/lim(x^2)
    we know that limcos(x) as x approaches 0 is 1, so we end up:
    lim((e^(x^2)-1)/x^2), which is equal to 1. Could anyone explain where I'm wrong on my second method?

    Thanks

    You cannot use arithmetic of limits if you don't know that ALL the limits involved exist finitely

    In the present case, \lim_{x\to 0}\frac{e^{x^2}-\cos x}{x^2}\neq \frac{\lim_{x\to 0}e^{x^2}-\lim_{x\to 0}\cos x}{\lim_{x\to 0}x^2} since the limit in the DENOMINATOR equals zero...!

    And after this you allow yourself to calculate ONLY the limit of the cosine but NOT the other ones....wrong.

    Perhaps the simplest way is using L'Hospital's rule:

    \lim_{x\to 0}\frac{e^{x^2}-\cos x}{x^2}=\lim_{x\to 0}\frac{2xe^{x^2}+\sin x}{2x}=\lim_{x\to 0}\left[e^{x^2}+\frac{1}{2}\,\frac{\sin x}{x}\right] = 1 + \frac{1}{2}=\frac{3}{2}

    Tonio
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  5. #5
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    hey tonio, thanks for your reply

    what I mean is:
    lim(e^(x^2))-lim(cos(x)). Since lim(cos(x))=1, we replace 1 with lim(1) (as it is also equal to 1), so we have:

    lim(e^(x^2))-lim(1), which should be equal to lim(e^(x^2)-1)

    :|
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    ps: i think this should be listed at the pre-calculus section, sorry abt it
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  7. #7
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    Quote Originally Posted by FreeedomOverdose View Post
    hey tonio, thanks for your reply

    what I mean is:
    lim(e^(x^2))-lim(cos(x)). Since lim(cos(x))=1, we replace 1 with lim(1) (as it is also equal to 1), so we have:

    lim(e^(x^2))-lim(1), which should be equal to lim(e^(x^2)-1)

    :|


    I know that's what you meant, and if you hadn't what you have in the denominator it'd be correct, but then do it all at once: since \lim_{x\to 0}e^{x^2}=1 , substitute both limits by their numerical value and then you get 0 ...but then you don't get what you wanted to, right?

    tonio
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  8. #8
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    so, the denominator is the problem?! So am I correct to say that:

    lim(e^x^2-cos(x))=lim(e^x^2-1)?!
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  9. #9
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    Quote Originally Posted by FreeedomOverdose View Post
    so, the denominator is the problem?! So am I correct to say that:

    lim(e^x^2-cos(x))=lim(e^x^2-1)?!
    If \lim_{x\to a}f(x),\lim_{x\to a}f(x) exist then \lim_{x\to a}\left\{f(x)\pm g(x)\right\}=\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)
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  10. #10
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    thank you guys! Everything is clear now. I just got myself into a vicious circle, happens sometimes :P.
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