2

$$\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2}$$

Attempt:

According to me, it has to be $1$, since the outermost exponent tends to $0$.

But anyways, would taking the limit as equal to $L$. And taking ($\log$) on both sides help?

Hanul Jeon
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Shashank Holla
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  • The use of $\sin^2 x$ in this expression seems strange. You could replace $y = \sin^2 x$ and look at the limit as $y \to 0_+$; I'm pretty sure it would be the same. – Michael Seifert Dec 08 '16 at 17:02
  • It would still be a $ (1^{1/y}+... n^{1/y})^y $ limit. Where can I go from here? – Shashank Holla Dec 08 '16 at 17:05
  • Also, are you sure that this equation is correct? Plotting the left-hand side as a function of $x$ (or $y$) seems to imply that the limit is $n$, not $n(n+1)/2$. – Michael Seifert Dec 08 '16 at 17:08
  • Yes. I am positive it is correct since I saw this on a worksheet I found online. Can you tell me how did you plot it? – Shashank Holla Dec 08 '16 at 17:15
  • Here it is on Wolfram Alpha for n = 4. Here it is for n = 10. These aren't proofs, of course, but they're perhaps indicative that there's a problem with the equation as stated. – Michael Seifert Dec 08 '16 at 17:23
  • @Shashank, I get the same results as Michael Seifert by plotting the function for various values of $n$ in Maple, using the sum function. It's not because you saw a result on line that it is true. – MasB Dec 08 '16 at 17:24
  • "I am positive it is correct since I saw this on a worksheet I found online" This sentence is truly a gem. @Shashank: Do you realize one finds both excellent stuff and pure junk "online"? What is the site you found this on? – Did Dec 08 '16 at 18:26
  • @Did Absolutely. Why I said it is correct because this has been solved before. Probably the answer key was wrong – Shashank Holla Dec 09 '16 at 02:32
  • @MichaelSeifert The approximation does tend to n. Looks like the problem is wrong. Should I update the question? – Shashank Holla Dec 09 '16 at 02:33
  • I don't think it's strictly necessary to edit the question. But if you do decide to do so, be sure to leave the original equation in it (noting that it's incorrect) so that future readers won't be confused about what was going on. And, of course, be sure to accept whichever answer below is the best one. – Michael Seifert Dec 09 '16 at 03:01
  • What is the site you found this on? – Did Dec 09 '16 at 06:26
  • @Did Couldnt find the link. So uploaded the worksheet for you. https://drive.google.com/file/d/0Bx4WpEnKvEioQW54ZXM4aF95NWc/view?usp=sharing – Shashank Holla Dec 09 '16 at 15:18

2 Answers2

3

Let us consider $n=2$ first, for simplicity, and use $x^2$ instead of $\sin^2x$, as $x\to0$. As you suggested, since the function has the indeterminate form $\infty^0$ as $x\to0$, let us consider the logarithm of the original function $$ x^2\log\left(1+2^{1/x^2}\right) = \frac{\log\left(1+2^{1/x^2}\right)}{1/x^2}, $$ which, as $x\to 0$, is of the form $\infty/\infty$. We can apply l'Hospital's rule getting $$ \frac{\log 2}{1+2^{-1/x^2}}, $$ which tends to $\log 2$ as $x\to0$. Therefore $$ \lim_{x\to 0} \left(1+2^{1/x^2}\right)^{x^2}=2. $$ In general, we may apply l'Hospital's rule to $$ \frac{\log\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)}{1/x^2} $$ getting $$ \frac{\left(\frac{2}{n}\right)^{1/x^2}\log 2+\left(\frac{3}{n}\right)^{1/x^2}\log 3+\ldots+\log n}{1+\ldots + \left(\frac{3}{n}\right)^{1/x^2}+\left(\frac{2}{n}\right)^{1/x^2}+n^{-1/x^2}}, $$ which tends to $\log n$ as $x\to0$, proiving $$ \lim_{x\to0}\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)^{x^2}=n. $$

Brightsun
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2

For any $n$ numbers $a_1, \ldots, a_n > 0$, we have

$$ \begin{align} \lim_{p\to+\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} &= \max\{ a_1, \ldots, a_n \}\tag{*1a}\\ \lim_{p\to-\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} &= \min\{ a_1, \ldots, a_n \}\tag{*1b} \end{align}$$

I will only justify $(*1a)$. Relabeling the numbers if necessary, we only need to study the case $0 < a_1 \le a_2 \ldots \le a_n$. For any $p > 0$, we have

$$ a_n = (a_n^p)^{1/p} \le (a_1^p + \ldots + a_n^p)^{1/p} \le (a_n^p + \ldots + a_n^p)^{1/p} = n^{1/p} a_n$$ Since $\lim\limits_{p\to+\infty} n^{1/p} = 1$, by squeezing, $(*1a)$ follows.

Apply this to the case $(a_1,\ldots,a_n) = (1,\ldots,n)$ and notice as $x \to 0$, $\frac{1}{\sin^2x} \to +\infty$, we get

$$\lim_{x\to 0} \left(1^{1/\sin^2 x} + \ldots + n^{1/\sin^2 x}\right)^{\sin^2 x} = \max\{ 1, \ldots, n \} = n$$

achille hui
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