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I was reading this post and I don't understand why I can't do this:

\begin{align*} \lim_{x \to \infty} \frac{1^{99} + 2^{99} + \cdots + x^{99}}{x^{100}} &= \lim_{x \to \infty} \frac{1^{99}}{x^{100}} + \lim_{x \to \infty} \frac{2^{99}}{x^{100}} + \cdots + \lim_{x \to \infty} \frac{x^{99}}{x^{100}} \\ &= 0 + 0 + \cdots + \lim_{x \to \infty} \frac{1}{x} \\ &= 0 \end{align*}

I know that isn't the correct answer but I want to know why it's not. Why does it fail?

Community
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francolino
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1 Answers1

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It fails because you're incorrectly using the linearity of limits. You can split a limit over a fixed sum, but the sum you're examining expands with $x$. Just like Riemann sum does (hint, hint!).

Theo Bendit
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  • What do you mean by "the sum you're examining expands with $x$"? – francolino Jul 02 '15 at 19:54
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    I mean that the number of terms in the sum depends on $x$. You can distribute the limit over a sum with a fixed number of terms, not over a sum where the number of terms increases with $x$. The classic example of where this doesn't work is Riemann sums: you are looking at a sum of areas of rectangles, each of which individually go to $0$, but overall, the limit of the Riemann sums will not necessarily be $0$. – Theo Bendit Jul 02 '15 at 19:58
  • Oh, I see it now. Thank you a lot. :) – francolino Jul 02 '15 at 19:59
  • http://math.stackexchange.com/questions/1280838/finding-limits-using-definite-integrals-lim-n-to-infty-sumn-k-1-frack4/1280865?noredirect=1#comment2741297_1280865 – Idris Addou Jul 03 '15 at 18:34