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How to calculate

\begin{align} \lim_{x\to +\infty} \frac{x^{\sin(x)+1}}{x} \end{align}

By looking at $x^{1+\sin(x)}$ plot I can see that since it oscillates between $1$ and $x^2$ the limit does not exist. So my main question is:

If the limit does not exist then what is the growth relationship between the two functions?

And my minor question is: how to prove that this limit does not exist?

KFkf
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  • Since $\sin{x}$ oscillates, the limit you want to compute does not exist, i.e. it does not converge to a single value. – Ritz Sep 17 '14 at 14:20
  • so what about the growth relationship between two functions ? – KFkf Sep 17 '14 at 15:06
  • What do you mean by the term growth relationship? – Ritz Sep 17 '14 at 15:36
  • do they have the same growth speed or is one of them faster than the other? – KFkf Sep 17 '14 at 19:32
  • To me this question makes no sense. Naturally, the denominator grows as $O(x)$. The numerator varies from $O(x)$ as $\sin(x) = 1$ or $O(1/x)$ as $\sin(x) = -1$. – Ritz Sep 18 '14 at 08:27

1 Answers1

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Use the sequence $x_n=n\pi$ which gives $$\frac{(x_n)^{\sin x_n}}{x_n}=\frac1{x_n}\underset{n\to\infty}\rightarrow0$$ and the sequence $y_n=\frac\pi2+2n\pi$ that gives $$\frac{(y_n)^{\sin y_n}}{y_n}=1\underset{n\to\infty}\rightarrow1.$$ You have found two sequences going to infinity that have different limits. Therefore, there is no limit for $x^{\sin x}/x$ when $x\to\infty$.

Tom-Tom
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