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Math Help - Limits.

  1. #1
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    Limits.

    "Find the limit or explain why it does not exist"
    Limits.-untitled.jpg
    I'm having trouble with these problems, I can't factor/these into a different form.
    Having trouble with the algebra.
    I've just started calculus and we haven't gone very far so we've just covered limit laws.

    I do understand for number 7 that "lim sin(x)/x = 1", but lim sin(2x)/(2x + 3x) doesn't seem right.
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  2. #2
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    Re: Limits.

    Quote Originally Posted by Zirrick View Post
    "Find the limit or explain why it does not exist"
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    I'm having trouble with these problems, I can't factor/these into a different form.
    Having trouble with the algebra.
    I've just started calculus and we haven't gone very far so we've just covered limit laws.

    I do understand for number 7 that "lim sin(x)/x = 1", but lim sin(2x)/(2x + 3x) doesn't seem right.
    5. Get a common denominator so that you can perform the subtraction, then divide by x. You should be able to substitute x = 1 then.

    7.
    \displaystyle \begin{align*} \frac{\sin{2x}}{5x} &= \frac{1}{5}\left(\frac{\sin{2x}}{x}\right) \\ &= \frac{2}{5}\left(\frac{\sin{2x}}{2x}\right) \end{align*}

    Now make use of the fact that \displaystyle \begin{align*} \lim_{X \to 0}\frac{\sin{X}}{X} = 1 \end{align*}.

    8. This is a perfect candidate for L'Hospital's Rule.

    9. Since logarithms are only defined for positive values of the independent variable, is it possible to approach 0 from the left? What does this tell you about this problem?

    11. Divide top and bottom by \displaystyle \begin{align*} x^3 \end{align*}.
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  3. #3
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    Re: Limits.

    For #7:

    limit sin(2x)/(5x) = limit sin(2x)/2x * limit 2x/5x = 1 * 2/5 = 2/5

    Just picture it.. if the denominator were equal to (2x), the limit would be 1, but instead, it is larger than that by a scale of (5/2)... therefore you need to divide by that much extra, to get 2/5.

    For #8, the same principle applies actually. As x -> 0 in this case, you can treat tan() as sin(), because you are dividing sin() by a quantity that approaches 1, cos(x). So the answer is 2/Pi.

    For #5, multiply both the numerator and the denominator of the big fraction by 2(2+x). It should then simplify nicely.
    Last edited by SworD; September 4th 2012 at 06:27 PM.
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  4. #4
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    Re: Limits.

    Hello, Zirrick!

    Here is #8 without using L'Hopital . . .


    8.\;\lim_{x\to0}\frac{\tan(2x)}{\tan(\pi x)}

    \frac{\tan(2x)}{\tan(\pi x)} \;=\;\dfrac{\frac{\sin(2x)}{\cos(2x)}}{\frac{\sin( \pi x)}{\cos(\pi x)}} \;=\; \frac{\sin(2x)}{1}\cdot\frac{1}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}

    . . =\;{\color{red}\frac{2x}{2x}}\cdot\frac{\sin(2x)}{  1}\cdot {\color{red}\frac{\pi x}{\pi x}}\cdot\frac{1}{\sin(\pi x)} \cdot \frac{\cos(\pi x)}{\cos(2x)} \;=\;\frac{2x}{\pi x}\cdot\frac{\sin(2x)}{2x}\cdot\frac{\pi x}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}


    \lim_{x\to0}\left[\frac{2}{\pi}\cdot\frac{\sin(2x)}{2x}\cdot\frac{ \pi x}{\sin(\pi x)}\cdot\frac{\cos(\pi x)}{\cos(2x)}\right] \;=\; \frac{2}{\pi}\cdot1\cdot1\cdot\frac{1}{1} \;=\;\frac{2}{\pi}

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