Find asymptotically equivalent function $f\sim g$

Question

I'm trying to find a asymptotically equivalent function $g$ for $f(x)= \sin{(\frac{\pi}{6^x})} - \dfrac{1}{2}$, ie $f \sim g$.

Should I use $\sin{x}$ Taylor expansion or something else?

Here, $x$ approaches to $1$.

@ Jakobian : sorry, edited, approached to 1 – Metso Mar 12 '19 at 18:57 — Metso, Mar 12 '19 at 18:57
Yes, I think Taylor series would be the best in here. – Jakobian Mar 12 '19 at 18:58 — Jakobian, Mar 12 '19 at 18:58
Is it in a neighbour hood of $0$ or of $\infty$ of of $1$? – Bernard Mar 12 '19 at 19:04 — Bernard, Mar 12 '19 at 19:04
@Bernard: a neighbourhood of 1 – Metso Mar 12 '19 at 19:06 — Metso, Mar 12 '19 at 19:06

Jakobian · Accepted Answer · 2019-03-12T19:14:22.433

3

Since $$(\sin(\pi/6^x)-1/2)' = -\ln 6 \cdot\frac{\pi \cos(\pi/6^x)}{6^x} $$ Taylor formula gives us $$\sin(\pi/6^x)-1/2 = -\ln 6 \cdot\frac{\pi\sqrt{3}}{12}(x-1)+o(x-1) $$ so $$\sin(\pi/6^x)-1/2 \sim -\ln 6 \cdot\frac{\pi\sqrt{3}}{12}(x-1)$$ as $x\to 1$.

edited Mar 12 '19 at 19:14

answered Mar 12 '19 at 19:03

Jakobian

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@Jacobian: Thank you for yor answer. It's interesting that $1/2=\sin(\pi/6)$. Does it make sense? – Metso Mar 12 '19 at 19:10
You forgot a factor $\pi$ in your derivative. – Bernard Mar 12 '19 at 19:12
@Bernard Thank you kindly – Jakobian Mar 12 '19 at 19:14
@Metso What do you mean? – Jakobian Mar 12 '19 at 19:16
@Jacobian: I mean in this exercise we can rewrite $\sin(\pi/6^x)-\sin(\pi/6)$. But it does not help. Thank you. – Metso Mar 12 '19 at 19:18

Find asymptotically equivalent function $f\sim g$

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