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Math Help - How are these indeterminate ?

  1. #1
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    How are these indeterminate ?

    0 * infinity
    1^ infinity
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  2. #2
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    Re: How are these indeterminate ?

    Quote Originally Posted by pratique21 View Post
    0 * infinity
    1^ infinity
    I guess these are indeterminate in the sense that if \lim_{x\to\infty} f(x)=0 and \lim_{x\to\infty} g(x)=\infty, then \lim_{x\to\infty}f(x)g(x) can be any number. For example, for a>0, \lim_{x\to\infty}\frac{1}{x}\cdot ax=a.

    Similarly, \lim_{n\to\infty}\left(1+\frac{a}{n}\right)=1 and \lim_{n\to\infty}n=\infty, but \lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n=e^a.
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  3. #3
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    Re: How are these indeterminate ?

    Hello, pratique21!

    How are these indeterminate? . 0\cdot\infty,\;\;1^{\infty}

    I'll explain the first one . . .


    Consider the function: . f(x) \:=\:\sin x\cdot\frac{1}{x}\,\text{ as }x\to0.
    We see that: . \lim_{x\to0}\left(\sin x \cdot\frac{1}{x}\right) \:=\:0\cdot\infty

    But we have a theorem that states: . \lim_{x\to0}\frac{\sin x}{x} \:=\:1

    . . Hence, we could say: . 0\cdot\infty \:=\:1


    Now consider the function: . g(x) \:=\:\sin2x\cdot\frac{1}{x}\,\text{ as }x\to0.

    We see that: . \lim_{x\to0}\left(\sin2x\cdot\frac{1}{x}\right) \:=\:0\cdot\infty

    But: . \lim_{x\to0}\frac{\sin2x}{x} \;=\;\lim_{x\to0} \left(2\!\cdot\!\frac{\sin2x}{2x} \right) \;=\; 2\cdot\lim_{x\to0}\frac{\sin2x}{2x} \;=\; 2\cdot1 \;=\;2

    . . Hence, we could say: . 0\cdot\infty \:=\:2


    In general, we can show that 0\cdot\infty can equal any number.

    Therefore, 0\cdot\infty is indeterminate.

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  4. #4
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    Re: How are these indeterminate ?

    Thank you Emakarov and soroban. I can understand it now. So that is the also why Infinity - infinity is indeterminate as well- It can take any value.
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  5. #5
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    Re: How are these indeterminate ?

    @pratique, \infty - \infty is also indeterminate.
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