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Wolfram understand this expression , but i need to do the limit when n tends to infinity of that expression .

As you can see in the wolfram web itself, in the last link it fails to understand the query.

How can i do that?

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    $\begingroup$ At the bottom, there is an "expanded form" of that expression. You can directly read the limit off that. $\endgroup$ – Daniel Fischer Jul 3 '13 at 18:19
  • $\begingroup$ Do you want to make WolframAlpha understand it or you understand it? Which one is your main goal? The limit can be computed by hand. $\endgroup$ – OR. Jul 3 '13 at 18:19
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    $\begingroup$ This question is fully un- selfcontained. $\endgroup$ – Did Jul 3 '13 at 19:09
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As suggested in the comment, at the very bottom of the page (first link), you'll see "Expanded form". If you merely "click" on the expression, the page will open to this page, where you'll find the limit of the expression stated explicitly:

Here's the limit (scroll down on the linked page to see it):

$$\lim_{n\to\pm\infty}\dfrac{4+24n+26n^2}{2n^2}=\dfrac{26}{3}\approx8.66667$$

Note that once you know the "expanded form" (or closed form) of the sum:

$$S(n) = \frac{4 + 24n + 26n^2}{3n^2},$$

Finding $\lim\limits_{n \to \infty} S(n)$ is fairly straightforward: We can see this limit more readily if we simply divide the numerator and denominator by $n^2:$

$$\lim_{n \to \infty}S(n) = \large \lim_{n \to \infty}\frac{\frac{4}{n^2} + \frac{24}{n} + 26}{3} = \frac {26}{3}$$

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  • $\begingroup$ two answers ;-) =1 $\endgroup$ – Amzoti Jul 4 '13 at 0:38

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