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Thread: 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 $\displaystyle {\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}$

    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...

    $\displaystyle {\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; Apr 30th 2009 at 09:48 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 $\displaystyle \frac{\tan(2x)}{\tan(\pi x)}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    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 $\displaystyle {\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}$

    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...

    $\displaystyle {\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 $\displaystyle \left(\frac{2}{\pi}\right)\cdot\frac{\frac{\sin(2x )}{2x}}{\frac{\sin(\pi x)}{\pi x}}\to\frac{2}{\pi}$
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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 $\displaystyle x\to 0$?

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

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

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

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

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

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

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

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

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

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

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

    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.

    $\displaystyle {\sin(ax) \over \sin(b x)} \to {a\over b}$ 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: $\displaystyle \epsilon$ is generally used to represent an arbitrary distance, usually one that is very small. $\displaystyle d(p,q)$ is the distance (a real number) between $\displaystyle p$ and $\displaystyle q$.

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

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

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

    So intuitively, that's why the proof works.
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