0
$\begingroup$

Original limit: $\lim \limits_{n \to \infty} \frac{\sqrt[n]{n}}{\sqrt n}\sin(3+n^5)$

What is this limit? :$\lim \limits_{n \to \infty} \frac{\sqrt[n]{n}}{\sqrt n}$

  • 0
    What does $x$ have to do with anything?? – 2017-02-25
  • 0
    Wrote x by mistake and can't edit it to n – 2017-02-25
  • 0
    Of course you can edit, it's your question. I edited it for you this time. – 2017-02-25
  • 0
    @rubik I couldn't edit it, it said error – 2017-02-25
  • 1
    If you know [that $\lim\limits_{n\to\infty} \sqrt[n]n=1$](/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1), the rest should be easy. – 2017-02-25

1 Answers 1

6

Use the fact that $$\lim \limits_{n \to \infty}\sqrt[n]{n} = 1$$ It could be verified by rewriting $\lim \limits_{n \to \infty}\exp^{\frac1n\ln(n)} = \exp^{\lim_{n \to \infty}\frac1n\ln(n)} = \exp^{0} = 1.$

  • 1
    ... or simply becasue $a^n>n$ for $n\gg 0$ and any fixed $a>1$. – 2017-02-25
  • 0
    Using this I get that the original limit is 0 by sandwich theorem. Thanks for the help. – 2017-02-25