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I am wondering what is $\frac{0}{0}$ and $\frac{\infty}{\infty}$?

In my impression, both are undefined. But then I need to prove that

$$\lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt} = 1 \text{ when } x \leq 0,$$ where $g(x) = f(x)f(1-x)$, and

\begin{equation*} f(x) = \left\{ \begin{array}{ll} e^{-1/x^2} & x > 0\\ 0 & x \leq 0 \end{array} \right. \end{equation*}


My attempt:

I split the demominator and got 0/0: \begin{eqnarray} h(x) & =& \lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt}\\ &= & \lim_{n \rightarrow \infty}\frac{\int_{-n}^x f(t)f(1-t)dt}{\int_{-n}^x f(t)f(1-t)dt+\int_x^n f(t)f(1-t)dt}\\ & =& \lim_{n \rightarrow \infty}\frac{0}{0+\int_x^n f(t)f(1-t)dt} \end{eqnarray}

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    $\begingroup$ If $a_n \to \infty$, and $b_n \to \infty$, then $\frac{a_n}{b_n}$ can still have a well-defined (finite) limit. $\endgroup$ – Daniel Fischer Jun 28 '13 at 18:59
  • $\begingroup$ Thanks @DanielFischer. For this question, I split the demominator and got 0/0: $h(x) = \lim_{n \rightarrow \infty} \frac{\int_{-n}^x g(t)dt}{\int_{-n}^n g(t)dt} = \lim_{n \rightarrow \infty}\frac{\int_{-n}^x f(t)f(1-t)dt}{\int_{-n}^x f(t)f(1-t)dt+\int_x^n f(t)f(1-t)dt}$ $\endgroup$ – WishingFish Jun 28 '13 at 19:00
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    $\begingroup$ Well, $\int_x^n f(t)f(1-t)\,dt$ is a positive constant for large enough $n$ ($n \geqslant 1$). So you don't get $\frac00$. $\endgroup$ – Daniel Fischer Jun 28 '13 at 19:05
  • $\begingroup$ Oops!~ Yes I was asked to prove exactly what I found... I thought I was asked to prove it =1.... $\endgroup$ – WishingFish Jun 28 '13 at 19:09
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I don't see how you get $1$ as that limit. Here's how I see it.

$g(t)$ is nonzero only when $t \in [0,1]$. Thus

$$\lim_{n \to \infty} \int_{-n}^n dt\, g(t) = \int_0^1 dt \, e^{1/x^2} e^{1/(1-x)^2}$$

Also, when $x < 0$:

$$\lim_{n \to \infty} \int_{-n}^x dt\, g(t) = 0$$

because $g(t) = 0$ when $t \in [-n,x]$. The limit is then zero.

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    $\begingroup$ Any particular reason for the downvote? $\endgroup$ – Ron Gordon Jun 28 '13 at 19:16
  • $\begingroup$ Hmm.. I got a downvote too, for the question. Anyways, I upvoted you. $\endgroup$ – WishingFish Jun 28 '13 at 19:23
  • $\begingroup$ I deserve the downvote because I asked the wrong question, which the limit should be 0 rather than 1 as I claimed. But you corrected me.. $\endgroup$ – WishingFish Jun 28 '13 at 19:24
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    $\begingroup$ @user83036: these things sometimes happen. You take the good with the bad here. Anyway, downvotes, when used correctly, make this site what it is. But sometimes, like here, i do not understand what I did wrong. As for you, you do NOT deserve a downvote for asking a question. $\endgroup$ – Ron Gordon Jun 28 '13 at 19:25
  • $\begingroup$ You are very kind, thank you. $\endgroup$ – WishingFish Jun 28 '13 at 19:26

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