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Math Help - l`hospital

  1. #1
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    l`hospital

    lim x->oo (x^2)sin(1/x)
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  2. #2
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    Quote Originally Posted by jeph View Post
    lim x->oo (x^2)sin(1/x)
    Do not use it!

    -1 \leq \sin \frac{1}{x} \leq 1 for all x\not = 0

    Thus,

    -x^2 \leq x^2 \sin \frac{1}{x} \leq x^2

    Now, \lim_{x\to 0} -x^2  = \lim_{x\to 0}x^2 =0

    By squeeze theorem,
    \lim_{x\to 0} x^2 \sin \frac{1}{x} =0
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  3. #3
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    are you sure? o.o

    the answer my prof provided was oo
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  4. #4
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    Quote Originally Posted by jeph View Post
    are you sure? o.o

    the answer my prof provided was oo
    I did it for x\to 0 instead of x\to \infty.

    In that case,
    \lim_{x\to \infty} x^2\sin \frac{1}{x} = +\infty

    But we do not need to use that rule.

    Instead for sufficiently large x consider the reciprocal, i.e. \frac{1}{x^2\sin \frac{1}{x}}

    We can think of this reciprocal function as a composition f\circ g where g = \frac{1}{x} and f=\frac{x^2}{\sin x}. But at x=0 we define f to be zero. Why? Because \lim_{x\to 0}f = \lim_{x\to 0} \frac{x^2}{\sin x} = \lim_{x\to 0} x \cdot \frac{x}{\sin x} = 0 \cdot 1 =0. This makes f continous at zero.

    Now \lim_{x\to \infty} g = 0 and f is continous at 0. By the limit composition rule the limit \lim_{x\to 0}f\circ g = f(0) = 0.

    We have shown that \lim_{x\to +\infty} \frac{1}{x^2\sin \frac{1}{x}} =0. Which implies that its reciprocal is +\infty that is, \lim_{x\to +\infty} x^2 \cdot \sin \frac{1}{x} = \infty.
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