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I had the following question in an exam. I wasn't able to solve the problem and I don't understand the solution.

I am supposed to find the limit of the following sequence:

$$ \lim_{x\to 1} \frac{\ln(x) - \sin(\pi x)}{\sqrt{x -1}} $$

where x > 1.

Now the solution proceeded as follows

$$ \ln(x) = x - 1 + O(|x-1|^2) $$

which didn't make sense. I know that the taylor expansion of the natural logarithm is $ \ln(x+1) = x + O(x^2) $. But why did they add the -1 and why replace x with (x-1)?

They then set:

$$ \sin(\pi x) = -\pi (x-1) + O(|x-1|^2) $$ which again isn't very clear to me. The taylor expansion of sin(x) is: $ \sin(x) = x + O(x^3) $. So how did they get to this expansion and how should I proceed to reach the same results?

Thank you

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Since $\ln(x+1)=x+O(x^2)$, $\ln(x)=\ln\bigl((x-1)+1\bigr)=x-1+O\bigl((x-1)^2\bigr)$. And since $\sin(x)=x+O(x^2)$,\begin{align}\sin(\pi x)&=-\sin(\pi x-\pi)\\&=-\sin\bigl(\pi(x-1)\bigr)\\&=-\pi(x-1)+O\bigl((x-1)^2\bigr).\end{align}So\begin{align}\frac{\ln(x)-\sin(\pi x)}{\sqrt{x-1}}&=\sqrt{x-1}\frac{\ln(x)-\sin(\pi x)}{x-1}\\&\to0\times(1-\pi)\\&=0.\end{align}

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  • $\begingroup$ Thanks, that makes a lot of sense. But is this only applicable if we know that x>1? And if not, how should we proceed then? Also: Isn't sin(x) = x + O(x^3)? $\endgroup$
    – Marwan
    Jun 18, 2019 at 9:25
  • $\begingroup$ This can be done only if $x>1$, because if $x<1$ then $\sqrt{x-1}$ makes no sense. And, yes, $\sin(x)=x+O(x^3)$, but as far as what I was doing was concerned, knowing that $\sin(x)=x+O(x^2)$ was enough. $\endgroup$
    – José Carlos Santos
    Jun 18, 2019 at 10:05
  • $\begingroup$ Ok. So this is only because we had $ \sqrt{x-1} $ in the denominator? Without that, it wouldn't have made sense to use the same strategy? And what do you mean as far as what you were doing was concerned? Is it because $ O(n^2) \in O(n^3) $ ? $\endgroup$
    – Marwan
    Jun 18, 2019 at 10:49
  • $\begingroup$ The assertion $\sin(x)=x+O(x^3)$ is stronger than $\sin(x)=x+O(x^2)$, but I only needed the weaker one. $\endgroup$
    – José Carlos Santos
    Jun 18, 2019 at 11:05

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