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$F'(x)=\frac{1}{x^n +4}$, where $n>1$.

Prove $\lim_{x\rightarrow\infty}F(x)$ exists.


So my thought went to use the Second Fundamental Theorem of Calculus:

$F(x)=\int_a^xF'(t)dt=\int_a^x\frac{1}{t^n+4}dt$

Taking the limits to infinity of both sides...

$\lim_{x\rightarrow\infty}F(x)=\lim_{x\rightarrow\infty}\int_a^x\frac{1}{t^n+4}dt$

And then I can't see whats next without doing something I'm unsure of.

My next thought involved if $\lim_{x\rightarrow\infty}F'(x)=0\Rightarrow \lim_{x\rightarrow\infty}F(x)<\infty$; as that limit of F(x) would be a constant. I've however found that to be easily disproven by playing with trig functions.

Any hints? I still think the fact $\lim_{x\rightarrow\infty}F'(x)=0$ is something very useful here.


So how about this? $\lim_{x\rightarrow\infty}F(x)=\lim_{x\rightarrow\infty}\int_a^x\frac{1}{t^n+4}dt=\int_a^\infty\frac{1}{t^n+4}<\int_a^\infty\frac{1}{t^n}dt=\int_a^\infty{t^{-n}}dt=\lim_{b\rightarrow\infty}[\frac{t^{-n+1}}{-n+1}]|_a^b$

$=\lim_{b\rightarrow\infty}\frac{1}{(b^n-1)(1-n)}-\frac{1}{(a^n-1)(1-n)}=-\frac{1}{(a^n-1)(1-n)}$; noting of course that $n>1$.

At this point If I select some specific arbitrary $a$ I could show its finite by Improper Integral Comparison Test?

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  • $\begingroup$ I would use the limit comparison test for improper integrals, comparing with $\int_a^\infty t^{-p}\,dt$. Edit: Or in this particular case, just a straight comparison test would work. $\endgroup$ – Daniel Schepler Apr 28 '17 at 22:53
  • $\begingroup$ I think I see what you're saying, I added a short edit going through it to be sure. Thank you for the reply! $\endgroup$ – OGV Apr 28 '17 at 23:15
  • $\begingroup$ It looks reasonable - though there is one minor error I didn't notice before. The formula should actually be $F(x) = F(a) + \int_a^x F'(t)\,dt$. It doesn't change the argument significantly, though, to carry through the $F(a) + \cdot$. $\endgroup$ – Daniel Schepler Apr 28 '17 at 23:24
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For $n>1$ and enough great $t$, we have

$$0 <\frac {1}{t^n+1}<\frac {1}{t^n} $$

and $\int^{+\infty}\frac {dt}{t^n} $ converges.

thus $\int^{+\infty}\frac {dt}{t^n+1} $ converges too. This means that $$\lim_{x\to+\infty}\int_a^x\frac {dt}{t^n+1} $$ exists.

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