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Math Help - How to find lim as x -> inf. of sin(x) algebraically

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    How to find lim as x -> inf. of sin(x) algebraically

    Given: \lim_{x \to \infty } \sin x

    I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.
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    Re: How to find lim as x -> inf. of sin(x) algebraically

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    Re: How to find lim as x -> inf. of sin(x) algebraically

    Quote Originally Posted by PhizKid View Post
    Given: \lim_{x \to \infty } \sin x
    I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.
    For each positive integer n define
    {x_n} = \left\{ {\begin{array}{*{20}{c}}  {\frac{\pi }{2} + 2n\pi ,}&{\text{n even}} \\   {\frac{{3\pi }}{2} + 2n\pi ,}&{\text{n odd}} \end{array}} \right.
    Now \left| {\sin ({x_n}) - \sin ({x_{n + 1}})} \right| = 2. That shows oscillation behavior.
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    Re: How to find lim as x -> inf. of sin(x) algebraically

    It says that: "x > d implies that |f(x) - L| < e." If x is outside the boundaries of delta, doesn't that mean |f(x) - L| > e?
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    Re: How to find lim as x -> inf. of sin(x) algebraically

    Quote Originally Posted by PhizKid View Post
    It says that: "x > d implies that |f(x) - L| < e." If x is outside the boundaries of delta, doesn't that mean |f(x) - L| > e?
    Intuitively, if you are evaluating a limit as x approaches infinity, you are making x approach a very large value. So if there was a limit, then if you choose a very large value, say "d", then by making x greater than this, you should be guaranteed to have the difference between f(x) and L be within your tolerance, e.
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    Re: How to find lim as x -> inf. of sin(x) algebraically

    I thought 'd' was the boundaries around 'x', so if 'x' is approaching a large value, then wouldn't the boundaries 'd' around 'x' be slightly larger than 'x'?
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    Re: How to find lim as x -> inf. of sin(x) algebraically

    Quote Originally Posted by PhizKid View Post
    I thought 'd' was the boundaries around 'x', so if 'x' is approaching a large value, then wouldn't the boundaries 'd' around 'x' be slightly larger than 'x'?
    You don't put a boundary around x when making x approach infinity, only a lower boundary (that happens to be very large).
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