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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $a_n \to M$

  • 1
    $\begingroup$ Thank you so much $\endgroup$
    – Peter F.
    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 $1$.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.