# Evaluation of limit to infinity with nth-root

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?

• Isn't also an $n$ in the exponent of $x^2$? Oct 23 '19 at 15:44
• @Dr.SonnhardGraubner yes, sorry I will edit it Oct 23 '19 at 15:45
• Wouldn't that be $1$?, Or do you need it with respect to $x$? Oct 23 '19 at 15:50
• 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. Oct 23 '19 at 15:51
• 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 Oct 23 '19 at 15:58

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 $$a_n \to M$$

• Thank you so much Oct 23 '19 at 16:01

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 $$1$$.

• I disagree. Take $x=1.5$ and see what happens. Oct 23 '19 at 15:52
• Yes you are right, I forgot the 2 which divides $x^2$ Oct 23 '19 at 15:54
• @DinnoKoluh how it will change? Oct 23 '19 at 15:55
• @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. Oct 23 '19 at 16:01
• @DinnoKoluh Thanks, now I understand Oct 23 '19 at 16:02