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Math Help - If anyone can offer me some help...

  1. #1
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    If anyone can offer me some help...

    I would really appreciate if anyone could lend me a hand (and a brain) to solve some of these questions
    where something similar may appear in my upcoming final!

    -Without using l'hopital's rule find the limit of tan(2x)/tan((pi)x) (sorry, I dont know how to type the pi sign)

    -how can I prove that functions cannot have two different limits at the same point?

    -Find the limit of sec(x sec^2x - tan^2x - 1) as X --> 1 (this question is particularly killing me)

    -Without using derivatives show that the equation x^3 - 15x + 1 = 0 has three solutions in the interval ( -4, 4 )

    Thankyou very much for reading this!!! I hope I don't bother anybody!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Rewrite  {\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}

    SINCE, I am assuming x is heading towards zero and we know that cosine of zero is 1.

    Also I plan to use (sin x)/x goes to 1 as x goes to 0.

    Thus...

     {\sin(2x) \over \sin(\pi x)} = \biggl({2\over \pi}\biggr) {(\sin(2x))/(2x) \over (\sin(\pi x))/(\pi x)}\to  {2\over \pi} .

    x^3 - 15x + 1 = 0 is a continuous function. So find some x's where the function is negative, positive...
    By the intermediate value theorem (think, the price is right) the function must equal zero between those x values.
    Last edited by matheagle; April 30th 2009 at 10:48 PM.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by onmiverse View Post
    I would really appreciate if anyone could lend me a hand (and a brain) to solve some of these questions
    where something similar may appear in my upcoming final!

    -Without using l'hopital's rule find the limit of \frac{\tan(2x)}{\tan(\pi x)}

    -how can I prove that functions cannot have two different limits at the same point?

    -Find the limit of \sec(x \sec^2x - \tan^2x - 1) as x\to1 (this question is particularly killing me)

    -Without using derivatives show that the equation x^3 - 15x + 1 = 0 has three solutions in the interval ( -4, 4 )

    Thankyou very much for reading this!!! I hope I don't bother anybody!
    For the first question, is it as x\to 0?

    ------------------------------------

    For the second question, let \{x_n\} be a sequence that converges to x and x'. (We want to show that x=x'.) Because of convergence, \forall~\epsilon>0, \exists~ N such that n>N \implies d(x,x_n)<\frac{\epsilon}{2}, and \exists~ N' such that n>N' \implies d(x_n,x')<\frac{\epsilon}{2}.

    So if n>\max\{N,N'\}, by the triangle inequality, d(x,x')\leq d(x,x_n)+d(x_n,x')<\frac{\epsilon}{2}+\frac{\epsil  on}{2} = \epsilon. Since \epsilon was arbitrary, we can conclude that x=x'.

    ----------------------------------

    For the third question, you know this function is continuous at x=1 (because it is a composition of continuous functions), so you can just plug in 1 for x to get \sec( 1\cdot\sec^2(1) - \tan^2(1) - 1)=\sec(0)=1

    ------------------------------------

    For the fourth question, f(-4) = -3 and f(-3) = 19, so by the Intermediate Value Theorem \exists~x_1\in(-4,-3) such that f(x_1)=0.

    Similarly, f(1)=-13 so the IMT says that \exists~x_2\in(-3,1) such that f(x_2)=0.

    Similarly, f(4)=5 so the IMT says that \exists~x_3\in(1,4) such that f(x_3)=0.

    Thus there are 3 zeroes in (-4,4).

    --------------------------------

    I hope I was some help.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by matheagle View Post
    Rewrite  {\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}

    SINCE, I am assuming x is heading towards zero and we know that cosine of zero is 1.

    Also I plan to use (sin x)/x goes to 1 as x goes to 0.

    Thus...

     {\sin(2x) \over \sin(\pi x)} = \biggl({\pi\over 2}\biggr) {(\sin(2x))/(2x) \over (\sin(\pi x))/(\pi x)}\to  {\pi\over 2} .

    x^3 - 15x + 1 = 0 is a continuous function. So find some x's where the function is negative, positive...
    By the intermediate value theorem (think price is right) the function must equal zero between those x values.
    I think it should be \left(\frac{2}{\pi}\right)\cdot\frac{\frac{\sin(2x  )}{2x}}{\frac{\sin(\pi x)}{\pi x}}\to\frac{2}{\pi}
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  5. #5
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    thank you very much, but I still don't understand...

    Quote Originally Posted by redsoxfan325 View Post
    For the first question, is it as x\to 0?

    ------------------------------------

    For the second question, let \{x_n\} be a sequence that converges to x and x'. (We want to show that x=x'.) Because of convergence, \forall~\epsilon>0, \exists~ N such that n>N \implies d(x,x_n)<\frac{\epsilon}{2}, and \exists~ N' such that n>N' \implies d(x_n,x')<\frac{\epsilon}{2}.

    So if n>\max\{N,N'\}, by the triangle inequality, d(x,x')\leq d(x,x_n)+d(x_n,x')<\frac{\epsilon}{2}+\frac{\epsil  on}{2} = \epsilon. Since \epsilon was arbitrary, we can conclude that x=x'.

    ----------------------------------

    For the third question, you know this function is continuous at x=1 (because it is a composition of continuous functions), so you can just plug in 1 for x to get \sec( 1\cdot\sec^2(1) - \tan^2(1) - 1)=\sec(0)=1

    ------------------------------------

    For the fourth question, f(-4) = -3 and f(-3) = 19, so by the Intermediate Value Theorem \exists~x_1\in(-4,-3) such that f(x_1)=0.

    Similarly, f(1)=-13 so the IMT says that \exists~x_2\in(-3,1) such that f(x_2)=0.

    Similarly, f(4)=5 so the IMT says that \exists~x_3\in(1,4) such that f(x_3)=0.

    Thus there are 3 zeroes in (-4,4).

    --------------------------------

    I hope I was some help.
    Thank you really all of you guys who replied... but I don't get the second answer...if you could expand upon that I would greatly appreciated it...
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  6. #6
    MHF Contributor matheagle's Avatar
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    Yup, I had it up-side down.

     {\sin(ax) \over \sin(b x)} \to {a\over b} as x heads towards zero.
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by onmiverse View Post
    Thank you really all of you guys who replied... but I don't get the second answer...if you could expand upon that I would greatly appreciated it...
    Draw f(x) via just a few points.
    Find some where f is positive and some in your interval where f is negative.
    Since f is continuous, it must cross zero between those x's.
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  8. #8
    Super Member redsoxfan325's Avatar
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    I'm not sure what calculus class you're in, but the proof I gave you was one I learned in Analysis, which is essentially a course entirely devoted to proving all the theorems that make up calculus.

    Some notation: \epsilon is generally used to represent an arbitrary distance, usually one that is very small. d(p,q) is the distance (a real number) between p and q.

    The idea behind this proof is that for large enough n, the distance between any given point in the sequence and the point to which the sequence converges becomes very small. So if the sequence converges to x and x', then for large enough n, x_n gets very close to x and x'. So if for large n, d(x_n,x)\approx0 and d(x_n,x')\approx0, then it makes sense that d(x,x')\approx0 as well. If two points have zero distance between them, then they must be the same point. Thus x=x'.

    So intuitively, that's why the proof works.
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  9. #9
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    thank you once again

    I get it now! Once it was explained with words I could easilly understand, and again, I am ever thankful for your generosity...
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  10. #10
    MHF Contributor matheagle's Avatar
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    My guess is that this is a calc 1 or 2 student and he/she had no idea what an epsilon is, besides a frat symbol.

    Quote Originally Posted by redsoxfan325 View Post
    I'm not sure what calculus class you're in, but the proof I gave you was one I learned in Analysis, which is essentially a course entirely devoted to proving all the theorems that make up calculus.

    Some notation: \epsilon is generally used to represent an arbitrary distance, usually one that is very small. d(p,q) is the distance (a real number) between p and q.

    The idea behind this proof is that for large enough n, the distance between any given point in the sequence and the point to which the sequence converges becomes very small. So if the sequence converges to x and x', then for large enough n, x_n gets very close to x and x'. So if for large n, d(x_n,x)\approx0 and d(x_n,x')\approx0, then it makes sense that d(x,x')\approx0 as well. If two points have zero distance between them, then they must be the same point. Thus x=x'.

    So intuitively, that's why the proof works.
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