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I'm trying to find, for the following function: $$U(x) := 2\log(x) + \sin(\log(x))\ ,\ \ x > e$$ the limit: $$\lim_{t \rightarrow \infty}\dfrac{U(t x)}{U(t)}$$ My initial idea was to expand $\sin(\log(x))$ into its polynomial form but the result is a series of logs. The condition that $x>e$ seems to suggest that these logs are all greater than $1$ for any $\alpha >0$ and therefore non-negligible. But then, I am struggling to say what that limit is above because surely each successive power in the expansion contributes even more to the limit than the previous one?

Any help is appreciated. Thank you.

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    How about $x$ in the limit? Is it fixed? – 2017-02-21
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    Ah ok, that's different – 2017-02-21
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    Sorry I made a typo; I've relabelled things so that it makes more sense now. Yes I believe it is fixed. – 2017-02-21
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    Remark that $\sin$ is bounded. Moreover, write $\log(tx)=\log t+\log x$. – 2017-02-21
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    Thanks for the hint. I've posted an idea below too - not sure if it's right. – 2017-02-21

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For fixed $x>e$ we have

$U(tx)=2 \log(t)+2 \log(x)+ \sin(\log t+ \log x)$

Now consider

$\dfrac{U(t x)}{U(t)}=\dfrac{\frac{U(t x)}{\log t}}{\frac{U(t)}{\log t}}$

and show

$\lim_{t \rightarrow \infty}\dfrac{U(t x)}{U(t)}=2$

Edit: we have $\lim_{t \rightarrow \infty}\dfrac{U(t x)}{U(t)}=1$

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    Isn't the limit $1$? Don't we have: $\Bigl|\dfrac{U(tx)}{U(t)}\Bigl| \leq \dfrac{2(\log(t) + \log(x)) + 1}{2\log(t) - 1} = \dfrac{2(1 + \frac{\log(x)}{\log(t)}) + \frac{1}{\log(t)}}{2 - \frac{1}{\log(t)}}$ (and a similar expression for the other inequality) giving a limit of $\frac{2}{2} = 1$? – 2017-02-21
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    Yes, you are right ! – 2017-02-21