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Math Help - 2 more limits questions?

  1. #1
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    Nov 2009
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    87

    2 more limits questions?

    Question 1: Evaluate limx->infinity [(1) / (sin(1/x))]

    Question 2:
    a) Evaluate limx->infinity (1) / (e^x)
    b) Evaluate limx->negative infinity (1) / (e^x)

    Any help is appreciated.
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  2. #2
    Member
    Joined
    May 2010
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    Hi there. You should try using latex, its better for all of us using latex code. Its easier to read, and there are expressions which are really complicated to write the way you're writing it.

    \displaystyle\lim_{x \to{\infty}}{\displaystyle\frac{1}{\sin(\frac{1}{x  })}}

    So. What do you know about the sine? You know that -1\leq{\sin t}\leq{1}

    So: -1\leq{\sin(\frac{1}{x})}\leq{1}

    But, what happens as x tends to infinity? then \frac{1}{x} tends to zero. And as the \frac{1}{x} tends to zero the entire denominator tends to zero. And then what happens with the entire expression?

    And you know that \frac{1}{x}\neq{0}\forall{x\in{\mathbb{R}}}

    This is the way you solve it:

    \displaystyle\lim_{x \to{\infty}}{\displaystyle\frac{1}{\sin(\frac{1}{x  })}}=\displaystyle\frac{\displaystyle\lim_{x \to{\infty}}{1}}{\displaystyle\lim_{x \to{\infty}}{\sin(\frac{1}{x})}}

    \displaystyle\lim_{x \to{\infty}}{1}=1
    And
    {\displaystyle\lim_{x \to{\infty}}{\sin(\frac{1}{x}}=0

    From here: \displaystyle\frac{\displaystyle\lim_{x \to{\infty}}{1}}{\displaystyle\lim_{x \to{\infty}}{\sin(\frac{1}{x})}}=+\infty

    Try to solve the others on your own. If you've got any doubts just ask it. And remember, you're not evaluating the function at the point you evaluate the limit, you are evaluating the limit at which the function tends as it gets closer and closer to a point. So, you're never dividing by zero, and actually the function never gets to infinity, the infinity is just an idea, a concept, as the limits are. It means that the function gets bigger and bigger as x gets bigger and bigger, or as the sine gets closer and closer to zero.

    Bye there.
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