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Math Help - Infinite limits

  1. #1
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    Infinite limits

    Note: This is for a Calculus 1 course that has just started, so appropriately I can only use algebra/trig to solve these

    \lim_{x \to \infty } x \sin \frac{1}{x}

    I know sin(1/x) is always between 0 and 1, but sin(1/x) is getting close to 0 as x approaches infinity (sin of something close to 0 is close to 0), and so I'm getting infinity * 0 which doesn't really make any sense.

    How can I show my work algebraically?
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    Re: Infinite limits

    Quote Originally Posted by PhizKid View Post
    Note: This is for a Calculus 1 course that has just started, so appropriately I can only use algebra/trig to solve these

    \lim_{x \to \infty } x \sin \frac{1}{x}

    I know sin(1/x) is always between 0 and 1, but sin(1/x) is getting close to 0 as x approaches infinity (sin of something close to 0 is close to 0), and so I'm getting infinity * 0 which doesn't really make any sense.

    How can I show my work algebraically?
    Write it as \displaystyle \begin{align*} x\sin{\left(\frac{1}{x}\right)} = \frac{\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x}} \end{align*} then make the substitution \displaystyle \begin{align*} u = \frac{1}{x} \end{align*}. Note that as \displaystyle \begin{align*} x \to \infty, u \to 0 \end{align*}, giving

    \displaystyle \begin{align*} \lim_{x \to \infty} x\sin{\left(\frac{1}{x}\right)} &= \lim_{x \to \infty}\frac{\sin{\left(\frac{1}{x}\right)}}{\frac  {1}{x}} \\ &= \lim_{u \to 0}\frac{\sin{u}}{u} \\ &= 1 \end{align*}
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    Re: Infinite limits

    Ah, I used substitution again for 1/x. Multiplied num/denom by the sub, and a form of sinx/x appears which can be then cancelled out, and left with 1x/x = 1.
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    Re: Infinite limits

    I have a similar problem though, but not sure if I need to create a new thread:

    sinx/x as x approaches infinity

    if I use substitution again, and use 1/u for x, I get lim as (1/u) approaches 0, sin(1/u)/(1/u) which is just 1, but that's incorrect... (since, intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence)
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    Re: Infinite limits

    Quote Originally Posted by PhizKid View Post
    I have a similar problem though, but not sure if I need to create a new thread:

    sinx/x as x approaches infinity

    if I use substitution again, and use 1/u for x, I get lim as (1/u) approaches 0, sin(1/u)/(1/u) which is just 1, but that's incorrect... (since, intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence)
    The product of a bounded function with a function that approaches 0 is 0.
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    Re: Infinite limits

    So is that like saying sinx * (1/x) as x approaches infinity = sinx * 0?
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    Re: Infinite limits

    Quote Originally Posted by PhizKid View Post
    So is that like saying sinx * (1/x) as x approaches infinity = sinx * 0?
    Sort of.
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  8. #8
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    Re: Infinite limits

    So on an exam if I were asked to show the work, what would I put?
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    Re: Infinite limits

    Quote Originally Posted by PhizKid View Post
    So on an exam if I were asked to show the work, what would I put?
    Exactly what I put. \displaystyle \begin{align*} \sin{x} \end{align*} is a bounded function, and \displaystyle \begin{align*} \frac{1}{x} \to 0 \end{align*} as \displaystyle \begin{align*} x \to \infty \end{align*}, so the limit of their product is 0.
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