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Evaluate the limit:$$\lim_{x\to \infty}\frac{(x+1)^1+(x+2)^2+(x+3)^3+........(x+100)^{100}}{x^{10}+10^{10}}$$

In my book they taken the higher power out i mean $x^{100}$ out and then got the answer but my doubt is how can he take individual limits without knowing the continuity of the function. I got bit confused by this About the computation of the limit $ \lim_{x \to \infty} \frac{1^{99} + 2^{99} + \cdots + x^{99}}{x^{100}} $

So please explain.

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user5954246
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  • The denominator in you limit is always positive so the function is indeed continuous. – GaussTheBauss Feb 25 '16 at 17:08
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    There is no problem here because the number of terms in the sum does not depend on $x$. $\cfrac{(x+1)+\dots +(x+n)}{x}\underset{x\to\infty}{\longrightarrow}n$ but $\cfrac{(x+1)+\dots +(x+x)}{x}\underset{x\to\infty}{\longrightarrow}+\infty$. In fact, it's much easier to see by simplifying a bit first: $\cfrac{(x+1)+\dots +(x+n)}{x}\underset{x\to\infty}=\cfrac{nx+\frac{n(n+1)}{2}}{x}{\longrightarrow}n$ but $\cfrac{(x+1)+\dots +(x+x)}{x}=\cfrac{x+\frac{x(x+1)}{2}}{x}\underset{x\to\infty}{\longrightarrow}+\infty$ (where I used that $1+\dots + k = \cfrac{k(k+1)}{2}$). – xavierm02 Feb 25 '16 at 17:13
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    Should the denominator be $(x^{10}+10)^{10}$? In the current form the limit is trivially infinity. – Lutz Lehmann Feb 25 '16 at 17:16
  • The property you have about continuity is that if $f$ and $g$ are continuous, then $h:x\mapsto f(x)+g(x)$ is continuous (). By using that a finite number of times, you can know that if $f_1,\dots,f_n$ are continuous, then $h:x\mapsto f_1(x)+\dots+f_n(x)$ is continuous. What happens in the question you link is that you define $g:x\mapsto f_1(x)+\dots+f_x(x)$. So now you have to use () $x$ times to prove the continuity of $x$. But you see it's really weird to have the number of functions depend on the argument and that won't work. – xavierm02 Feb 25 '16 at 17:19
  • the link which i gave in that answer he got (the question asker) was not correct then how do i know this will work in my case as this didn't work in his case (i mean how i know my answer will be correct as questions are almost the same) – user5954246 Feb 25 '16 at 17:22
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    In your case, you have exactly $100$ terms in the sum. He had $x$, which causes the problem. – xavierm02 Feb 25 '16 at 17:24

1 Answers1

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Divide the numerator and the denominator by $x^{10}$, which is the highest power there.

Then look at the limit of the denominator as $x \to \infty$. That should be a number.

Next look at the limits of the first, second, ..., tenth term in the numerator. These are all numbers. Work out each term until you see the pattern. Pay special attention to the tenth term.

Then work out the eleventh term and a few of the following terms.

At that point you should see the answer.

Continuity is not an issue here. Since $x \to \infty$, you can always assume that $x > 1$ and everything is defined.

Hans Engler
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