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Math Help - Confirming a Limit Algebraically

  1. #1
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    Confirming a Limit Algebraically

    limx→0 sin x / 2x2-x

    I'm supposed to find the limit graphically (which I did; it's -1) and confirm it algebraically. I started by factoring the denominator 2x2-x to x(2x - 1). I'm trying to get rid of that x factor in the denominator but I have no clue what to do with the numerator sin x.
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    Re: Confirming a Limit Algebraically

    I suspect you are meant to write this as:

    $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{2x^2 - x} = \left(\lim_{x \to 0} \dfrac{\sin x}{x}\right)\cdot\left(\lim_{x \to 0} \dfrac{1}{2x - 1}\right)$

    Have you seen the limit of the factor on the left before?
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    Re: Confirming a Limit Algebraically

    I've seen in another problem that it is equal to 1, but I do not know why. Is that the most effective way of solving this?
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    Re: Confirming a Limit Algebraically

    Quote Originally Posted by Deveno View Post
    I suspect you are meant to write this as:

    $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{2x^2 - x} = \left(\lim_{x \to 0} \dfrac{\sin x}{x}\right)\cdot\left(\lim_{x \to 0} \dfrac{1}{2x - 1}\right)$

    Have you seen the limit of the factor on the left before?
    \displaystyle \lim_{x \to 0} \dfrac{\frac{d}{dx}\sin x}{\frac{d}{dx}2x^2 - x} = \lim_{x \to 0} \frac{\cos x}{4x-1}=-1
    Last edited by Jonroberts74; July 5th 2014 at 10:03 PM.
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    Re: Confirming a Limit Algebraically

    Using L'Hospital's Rule in this case isn't really valid, as finding the derivative of sin(x) requires already knowing that $\displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} = 1 \end{align*}$.
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    Re: Confirming a Limit Algebraically

    Confirming a Limit Algebraically-unit-circle.jpg

    This is a unit circle, obviously with a radius of 1 unit. The angle $\displaystyle \begin{align*} \theta \end{align*}$ is the angle swept out from the positive x axis in the anticlockwise direction, measured in radians. The green length is $\displaystyle \begin{align*} \cos{(\theta)} \end{align*}$, the red length is $\displaystyle \begin{align*} \sin{(\theta)} \end{align*}$ and the purple length is $\displaystyle \begin{align*} \tan{(\theta)} \end{align*}$. From the diagram, it's clear that the area of the segment is a little bigger than the area of the smaller triangle, and a little less than the area of the large triangle.

    We'll deal with positive angles for the moment:

    The area of the small triangle is $\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \end{align*}$, the area of the large triangle is $\displaystyle \begin{align*} \frac{1}{2}\tan{(\theta)} = \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \end{align*}$, and the area of the segment is $\displaystyle \begin{align*} \frac{\theta}{2\pi} \cdot \pi \cdot 1^2 = \frac{1}{2}\theta \end{align*}$, thus...

    $\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \leq \frac{1}{2}\theta &\leq \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \sin{(\theta)}\cos{(\theta)} \leq \theta &\leq \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \cos{(\theta)} \leq \frac{\theta}{\sin{(\theta)}} &\leq \frac{1}{\cos{(\theta)}} \\ \frac{1}{\cos{(\theta )}} \geq \frac{\sin{(\theta)}}{\theta} &\geq \cos{(\theta ) } \\ \cos{(\theta )} \leq \frac{\sin{(\theta)}}{\theta} &\leq \frac{1}{\cos{(\theta)} } \end{align*}$

    And now as $\displaystyle \begin{align*} \theta \to 0, \cos{(\theta)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} \to 1 \end{align*}$, thus $\displaystyle \begin{align*} \frac{\sin{(\theta)}}{\theta} \to 1 \end{align*}$ as $\displaystyle \begin{align*} \theta \to 0 \end{align*}$.


    Of course this only proves the right hand limit, where you are approaching 0 from positive values of $\displaystyle \begin{align*} \theta \end{align*}$, you need to prove the left hand limit as well, where you make $\displaystyle \begin{align*} \theta \to 0 \end{align*}$ from negative values as well. But the proof of that is almost identical
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    Re: Confirming a Limit Algebraically

    Quote Originally Posted by cdbowman42 View Post
    I've seen in another problem that it is equal to 1, but I do not know why. Is that the most effective way of solving this?
    I don't know which tools it is assumed you already know. If you are allowed to use the fact that:

    $\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1$

    then I would think that's the way to go.

    It's an IMPORTANT limit, because it's essentially what is used to find the derivative of the sine function. As with finding the derivative of the exponential function, there are certain "chicken-or-egg" questions that come up in PROVING what this limit is. The "standard" proof, is essentially what ProveIt posted, and uses a geometric argument.

    Even as innocent, and as self-explanatory as ProveIt's proof seems, there are some underlying basic assumptions we take for granted, the chief of which is:

    The area of a triangle with base $b$ and height $h$ is: $\dfrac{bh}{2}$.

    You've probably never seen a proof of this, not because the proof is difficult, but because defining what "area" is (or even "length of a side of a triangle"), is harder than it appears to be at first glance.

    Nevertheless, ProveIt's argument is likely the best you will be able to understand, until you have access to some fairly sophisticated mathematical methods. And it IS true (just trust me. *Evil laugh*).
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    Re: Confirming a Limit Algebraically

    How your prove " \lim_{x\to 0}\frac{sin(x)}{x}" depends, of course, upon exactly how you have defined sin(x).

    We can, for example, define sin(x) to be the power series x- x^3/3!x+ \cdot\cdot\cdot= \sum_{n=0}^\infnty \frac{x^{2n+1}}{(2n+1)!}. Then it is easy to see that \frac{sin(x)}{x}= \frac{x- x^3/6+\cdot\cdot\cdot}{x}= 1- x^2/6+ \cdot\cdot\cdot which clearly has limit 1 as x goes to 0.

    Or we can define sin(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 0, y'(0)= 1" while defining cos(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 1, y'(0)= 0".

    Using those definitions, if we let y= (sin(x))', the derivative of sin(x), then y'= ((sin(x))')'= (sin(x))''= -(sin(x)) so that y''= -(sin(x))'= -y. That is, this new function also satisfies the equation y''= -y. Further y(0)= (sin(x))' at 0 which is 1 and y'(x)= (sin(x))''= - sin(x) is 0 at x= 0. By "uniqueness" of the solutions of these initial value problems, then, y(x)= cos(x). That is (sin(x))'= cos(x) and, since we have got that derivative without using the limit, we can use L'Hopital's rule: \lim_{x\to 0}\frac{sin(x)}{x}= \lim_{x\to 0}\frac{cos(x)}{1}= 1.
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    Re: Confirming a Limit Algebraically

    Haha, thanks, I completely get the proof. I was just thrown off because that question came up in the book before anything was said about that special limx→0 (sin x)/x. That seems to be the most logical way to solve the problem. I was just trying to find out if their was any other, solely algebraic way to find the limit without knowing that limx→0 (sin x)/x = 1.
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    Re: Confirming a Limit Algebraically

    Quote Originally Posted by HallsofIvy View Post
    How your prove " \lim_{x\to 0}\frac{sin(x)}{x}" depends, of course, upon exactly how you have defined sin(x).

    We can, for example, define sin(x) to be the power series x- x^3/3!x+ \cdot\cdot\cdot= \sum_{n=0}^\infnty \frac{x^{2n+1}}{(2n+1)!}. Then it is easy to see that \frac{sin(x)}{x}= \frac{x- x^3/6+\cdot\cdot\cdot}{x}= 1- x^2/6+ \cdot\cdot\cdot which clearly has limit 1 as x goes to 0.

    Or we can define sin(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 0, y'(0)= 1" while defining cos(x) to be "the unique solution to the initial value problem y''= -y, y(0)= 1, y'(0)= 0".

    Using those definitions, if we let y= (sin(x))', the derivative of sin(x), then y'= ((sin(x))')'= (sin(x))''= -(sin(x)) so that y''= -(sin(x))'= -y. That is, this new function also satisfies the equation y''= -y. Further y(0)= (sin(x))' at 0 which is 1 and y'(x)= (sin(x))''= - sin(x) is 0 at x= 0. By "uniqueness" of the solutions of these initial value problems, then, y(x)= cos(x). That is (sin(x))'= cos(x) and, since we have got that derivative without using the limit, we can use L'Hopital's rule: \lim_{x\to 0}\frac{sin(x)}{x}= \lim_{x\to 0}\frac{cos(x)}{1}= 1.
    Yes, I alluded to these methods in my earlier post. However, to delay differentiation of trigonometric functions until after defining power series (and the necessary methods of proving convergence, and showing term-by-term differentiation is justified within the radius of convergence, or even that multiplying a power series by another function yields another power series which is also convergent) seems rather akin to explaining the last scene of Romeo and Juliet before the play starts.

    Defining the sine function as the unique solution to an IVP is even more unstisfactory, from a pedagogical point of view: the existence and uniqueness of such an IVP is a deep theorem, unlikely to ever be encountered by someone just beginning calculus.

    It seems probable, even, that the OP may not even have been exposed to "epsilon-delta" limit proofs, making your cart in San Francisco, and your horse in New York.

    From a logical standpoint, it may indeed be more satisfactory to introduce logically prior concepts before using ideas that require them. I have no problem with this, but in practice, this is almost never done. For if so, ZF set theory, and the Peano axioms would be taught to grade school kids, so that they might begin studying Dedekind cuts and sequences in high-school, and finally, at the collegiate level, be introduced to the notion of a continuous function of a real variable (perhaps there would be advanced placement classes in factoring polynomials).
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