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Specifically, I ran into trouble when trying to solve $$\lim\limits_{n \to \infty} \int_1^n x^{-n} \, dx.$$

The integral is very easy to solve but what next? The limit $\displaystyle\lim\limits_{n\to \infty} \frac{n^{-n+1}-1}{-n+1}$ looks like $\displaystyle\lim\limits \frac{\infty^{-\infty}}{-\infty}$ which I'm not sure how to solve. Can I apply L'Hopital Rule here?

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  • $\begingroup$ Can you solve the definite integral? $\endgroup$ Sep 26, 2014 at 0:45
  • $\begingroup$ Hi @Meelo sure the integral is very easy to solve but what next? the limit $\lim\limits_{n\to \infty} \frac{n^{-n+1}-1}{-n+1}$ looks like $\lim\limits \frac{\infty^\infty}{-\infty}$ which I'm not sure how to solve. $\endgroup$
    – MatheMagic
    Sep 26, 2014 at 1:00
  • $\begingroup$ Can I apply L'Hopital Rule here? $\endgroup$
    – MatheMagic
    Sep 26, 2014 at 1:07
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    $\begingroup$ You've made a mistake in retaining the most significant term in the numerator. For large $n$, the constant term $-1$ will subsume the small positive contribution $n^{-n+1}$. Therefore your limit should be like $\frac{-1}{-\infty}$, which is probably easier for you anyway. $\endgroup$
    – hardmath
    Jan 4, 2015 at 19:04

4 Answers 4

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Hint: Let $t=\dfrac1x$ , and use the fact that $\displaystyle\lim_{a\to\infty}x^a=0$ when $x\in(0,1)$.

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As you wrote, you get $$ \lim_{n\to \infty} \frac{n^{-n+1}-1}{-n+1}. $$ Note that there is a minus sign in the exponent, so this should not look like $\frac{\infty^\infty}{-\infty}$ but something simpler: just find the limits of the numerator and denominator separately, and you should see the result.

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  • $\begingroup$ Sure, and that's gonna reduce to $\lim\limits_{n\to \infty} n^{-n}$, which will then be solved given the hint from @Lucian. (I forgot to put the minus sign in front of the infinity power in the original post, now it's fixed.) $\endgroup$
    – MatheMagic
    Sep 26, 2014 at 18:52
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Exponentiation (of extended real numbers) is defined and continuous there, so you can just plug things in, using $(+\infty)^{-\infty} = 0$.

Similarly, $(+\infty)^{+\infty} = +\infty$

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inf^(-inf)/-inf = 1/inf*(-inf) = -1 / inf^2 --> 0

$$\frac{\infty^{-\infty}}{-\infty} = \frac1{\infty\cdot(-\infty)} = -\frac1{\infty^2} \to 0$$

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  • $\begingroup$ When supplying an Answer three months after a previously answered Question, haste is not much of a benefit. The MathJax software here allows you to post in mathematical formulas and symbols. $\endgroup$
    – hardmath
    Jan 4, 2015 at 19:09
  • $\begingroup$ @user204751 It is really difficult to say what you wanted to write. I have edit your post; from my edit you should be able to see at least how mathematical expressions can be formatted and edit it further, if needed. For more information about writing math at this site see e.g. here, here, here and here. $\endgroup$ Jan 24, 2015 at 15:04
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    $\begingroup$ @MartinSleziak This user is unregistered and has not been seen since the day this was posted. It's probably a lost account, and will never return. Probably best to just judge whether it should be kept at all. $\endgroup$
    – user642796
    Jan 24, 2015 at 16:10

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