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Hey I am supposed to evaluate the limit:

$$\lim_{n\rightarrow \propto }\sqrt[n]{1+x^{n}+\left ( \frac{x^{2}}{2} \right )^{n}}, x\geq 0$$

My idea was that I can rewrite it in: $$e^{\lim_{n\rightarrow \propto }\left ( ln\left ( 1+x^{n}+\left ( \frac{x^{2}}{2} \right )^{n}\right )\right )} $$

But I do not know, what to do next.

Can anyone help me?

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  • $\begingroup$ Isn't also an $n$ in the exponent of $x^2$? $\endgroup$ Oct 23 '19 at 15:44
  • $\begingroup$ @Dr.SonnhardGraubner yes, sorry I will edit it $\endgroup$
    – Peter F.
    Oct 23 '19 at 15:45
  • $\begingroup$ Wouldn't that be $1$?, Or do you need it with respect to $x$? $\endgroup$
    – user712576
    Oct 23 '19 at 15:50
  • $\begingroup$ I think this depends on the value of $x$ since this will establish whether $x^n$ will dominate or $(\frac{x^2}{2})^n$ will dominate. $\endgroup$ Oct 23 '19 at 15:51
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    $\begingroup$ There are a few cases. $x < 1$: Then $x^n$ and $(x^2/2)^n$ will approach 0 and thus you will probably get a 1. If x>1 but x<2 then the $x^n$ term will be greater than 1, but the $x^2/2$ will be less than 1 and thus approach 0, thus giving x. Finally if $x>2$ then the $x^2/2$ term will dominate and you will get that result $\endgroup$ Oct 23 '19 at 15:58
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Denote $a_n$ the sequence and let $x \geq 0$ and $M:=\max\{1,x,\frac{x^2}{2}\}$

$$M \leq a_n \leq\sqrt[n]{3}M$$

So $a_n \to M$

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    $\begingroup$ Thank you so much $\endgroup$
    – Peter F.
    Oct 23 '19 at 16:01
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If $x>1$ then the term $\frac{x^2}{2}$ will only persist as $n\rightarrow \infty$ if the condition $\frac{x^2}{2}> x$ is satisfied and the solution will be $\frac{x^2}{2}$ if on the other hand we have that $\frac{x^2}{2} < x$ then only $x$ will persist and the solution is $x$. If $x\leq 1$ after taking the root the solution will be just $1$.

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    $\begingroup$ I disagree. Take $x=1.5$ and see what happens. $\endgroup$ Oct 23 '19 at 15:52
  • $\begingroup$ Yes you are right, I forgot the 2 which divides $x^2$ $\endgroup$ Oct 23 '19 at 15:54
  • $\begingroup$ @DinnoKoluh how it will change? $\endgroup$
    – Peter F.
    Oct 23 '19 at 15:55
  • $\begingroup$ @PeterF. I forgot the 2 in the denominator because as stated for $x = 1.5$, $\frac{1.5^2}{2}<1.5$ so only $1.5$ remains. $\endgroup$ Oct 23 '19 at 16:01
  • $\begingroup$ @DinnoKoluh Thanks, now I understand $\endgroup$
    – Peter F.
    Oct 23 '19 at 16:02

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