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Math Help - Sequences-Original Diverges, Inverse Converges?

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    Sequences-Original Diverges, Inverse Converges?

    Suppose f satisfies \lim_{x->0^+}f(x)=+\infty. Is it possible the sequence { f(\frac{1}{n})} converges? Explain.
    The answer to this question is "yes" (I don't know why), because the second part asks:
    Find a function f that
    supports this theory.
    I don't really understand why the answer is yes. The only function I can think of that goes to infinty as x goes to 0 is \frac{1}{x^n}.
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    Quote Originally Posted by WhoCares357 View Post
    The answer to this question is "yes" (I don't know why), because the second part asks:


    I don't really understand why the answer is yes. The only function I can think of that goes to infinty as x goes to 0 is \frac{1}{x^n}.

    Great. So you found a function, f(x)=\frac{1}{x^n} s.t \lim_{x\rightarrow 0^+}f(x)=\infty\,\,and\,\,f\left(\frac{1}{n}\right  )=n^n converges (and to zero)...so what's the problem, again?

    Tonio
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    Quote Originally Posted by tonio View Post
    Great. So you found a function, f(x)=\frac{1}{x^n} s.t \lim_{x\rightarrow 0^+}f(x)=\infty\,\,and\,\,f\left(\frac{1}{n}\right  )=n^n converges (and to zero)...so what's the problem, again?

    Tonio
    Oh. Haha. I was thinking of setting to x->infinty instead of x->0. Stupid me.
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