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Find the limit $$L=\lim_{x\to0}\dfrac{\ln(2x^2-2x+1)+\arcsin{2x}}{(\sqrt{1-4x}-1)(1-\cos(2x))}$$

First let's get rid of the square root.

$$L=\lim_{x\to0}\dfrac{\left[\ln(2x^2-2x+1)+\arcsin(2x)\right]\left(\sqrt{1-4x}+1\right)}{(1-4x-1)(1-\cos(2x))}$$

I thought it would be a good idea to write the limit as a sum of the following limits $$L\color{red}{\stackrel{?}{\color{red}{=}}}-\dfrac12\lim_{x\to0}\dfrac{\ln(2x^2-2x+1)}{x(1-\cos(2x))}-\dfrac12\dfrac{\arcsin(2x)}{x(1-\cos(2x))}=-\dfrac12L_1-\dfrac12L_2$$ Note that the first limit is $-\infty$ though, so the equality does not actually hold: $$L_1= \lim_{x\to0}\dfrac{\ln(2x^2-2x+1)}{x(1-\cos(2x))}\cdot\dfrac{2x^2-2x}{2x^2-2x}=2\lim_{x\to0}\dfrac{x-1}{1-\cos(2x)}\\=\lim_{x\to0}\dfrac{x-1}{\sin^2x}\cdot\dfrac{x^2}{x^2}=-\infty.$$ As a matter of fact, because of the negative coefficient $\left(-\dfrac12\right),$ it would become $+\infty$. This does not change anything of course. I just decided to mention it.

I don't see how to approach the problem if we are not supposed to somehow "split" the limit.

I'm looking for a solution that doesn't rely on any fancy split ideas, but rather on a general approach and intuitive ideas. Thanks!

Math Student
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  • If you want to split via a term ensure first that the limit of that term exists finitely. And if you wish to split via a factor then ensure that the limit of that factor exists finitely and is non-zero. – Paramanand Singh Feb 02 '24 at 01:18

3 Answers3

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As $u\to0$, $\sqrt{1+u}=1+\frac u2+o(u)$ and $\cos u=1-\frac{u^2}2+o(u^2)$ hence as $x\to0$, $$(\sqrt{1-4x}-1)(1-\cos(2x))\sim(-2x)(2x^2)=-4x^3.$$ This incites us to expand the numerator at the order $3$. As $u\to0$, $$\begin{align} \ln(1+u)&=u-\frac{u^2}2+\frac{u^3}3+o(u^3),\\ \arcsin u&=u+\frac{u^3}6+o(u^3), \end{align}$$ hence as $x\to0$, $$\begin{align} \ln(1-2x+2x^2)&=(-2x+2x^2)-\frac{4x^2-8x^3}2+\frac{(-2x)^3}3+o(x^3)\\ &=-2x+\frac{4x^3}3+o(x^3),\\ \arcsin(2x)&=2x+\frac{4x^3}3+o(x^3) \end{align}$$ and therefore, $$\lim_{x\to0} \frac{\ln(2x^2-2x+1)+\arcsin{2x}}{(\sqrt{1-4x}-1)(1-\cos(2x))}=\lim_{x\to0}\frac{\frac{8x^3}3}{-4x^3}=-\frac23.$$

Anne Bauval
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  • Why the downvote, please? – Anne Bauval Jan 30 '24 at 22:01
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    SE being SE users. +1 from me. – Math Student Jan 30 '24 at 22:02
  • Not clear why, maybe for some typos now fixed. – user Jan 30 '24 at 22:07
  • I like it. I would mention that in this type of problems one should expand the denominator first, to check the order of the expansion in the numerator. In this case $x\cdot x^2=x^3$ – Andrei Jan 30 '24 at 22:07
  • I agree, I shall reverse the order of some lines (blindly since at the moment, my browser doesn't want to typeset, whence the temporary typos). – Anne Bauval Jan 30 '24 at 22:10
  • @Andrei, may you clarify what do you mean by "expand the denomiranotr to check the order of the expansion in the numerator"? I am new to Taylor series, so such hints are really helpful for me to learn how to solve such problems. – Math Student Jan 30 '24 at 22:15
  • May I also ask how am I supposed to find the Taylor series of $\arcsin(x)$? I know how to derive the others, but not this one. – Math Student Jan 30 '24 at 22:19
  • Isn't it $8x^2-8x^3$ in the numerator when you plug $x=2x^2-2x$ into the expansion of $\ln(1+x)$? – Math Student Jan 30 '24 at 22:42
  • See Clever derivation of $\arcsin(x)$ Taylor series, especially the first comment by @RobertIsrael. But if you just want it at the order 3 like here, the simplest way is to plug $y=\arcsin x=x+ax^3+o(x^3)$ into $x=\sin y=y-\frac{y^3}6+o(y^3)$ (or vice-versa), to find $a$. – Anne Bauval Jan 30 '24 at 22:42
  • Also isn't the next term $\dfrac{8x^3(1-x)^3}{3}$? Where do these multipliers go? – Math Student Jan 30 '24 at 22:43
  • @AnneBauval, am I missing something, or there's an issue with the calculations for the Taylor series of $\ln(2x^2-2x+1)$? – Math Student Jan 30 '24 at 22:47
  • If $u=-2x+2x^2$ then $u^2=4x^2-8x^3+4x^4=4x^2-8x^3+o(x^3)$, and $u^3=-8x^3(1-x)^3=-8x^3+o(x^3)$. – Anne Bauval Jan 30 '24 at 22:47
  • @AnneBauval, oh, I see. May I ask you what's the intuition behind this $o(f(x))$ symbol? – Math Student Jan 30 '24 at 22:48
  • https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation $o(x^3)$ means "some function $f(x)$ which tends to $0$ faster than $x^3$", i.e. $\lim_{x\to0}f(x)/x^3=0$. – Anne Bauval Jan 30 '24 at 22:51
  • @MathStudent You know that the denominator is $0$ at $x=0$. That means that the dominant term is $x$ to some power, that you want to find out. You already did half of the expansion, when you found in the first parenthesis $1-4x-1=-4x$. The second parenthesis is $1-cos(2x)\approx 1-1+2x^2=2x^2$. So the dominant term is $x^3$ – Andrei Jan 30 '24 at 23:22
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Referring to standard limits and the other known results, we recognize that the denominator is $\sim 4x^3$, therefore

$$\dfrac{\ln(2x^2-2x+1)+\arcsin{2x}}{(\sqrt{1-4x}-1)(1-\cos(2x)}=\dfrac{\ln(2x^2-2x+1)+\arcsin{2x}}{4x^3}\dfrac{4x^3}{(\sqrt{1-4x}-1)(1-\cos(2x)}$$

with (the first term is related to the definition of derivative)

$$\dfrac{(\sqrt{1-4x}-1)(1-\cos(2x))}{4x^3}=\dfrac{\sqrt{1-4x}-1}{x}\dfrac{1-\cos(2x)}{4x^2} \to -2 \cdot \frac 12=-1$$

and

$$\dfrac{\ln(2x^2-2x+1)+\arcsin{2x}}{4x^3}=\dfrac{\ln(2x^2-2x+1)+2x}{4x^3}+2\dfrac{\arcsin{2x}-2x}{8x^3} \to \frac13+\frac13 =\frac 23$$

and all boils down in the first one which can be proved by l'Hopital (one step) or Taylor expansion and which is related to

$$\lim_{x\to 0} \frac{\log(1+x)-\left(x-\frac{x^2}2\right)}{x^3} \to \frac13$$

user
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Instead of doing the split in terms of $L_1,L_2$ it is better to continue with further algebraic manipulation. Thus we can start from the expression $$-\frac{1}{2}\cdot\frac{\log(2x^2-2x+1)+\arcsin 2x}{x(1-\cos 2x)}$$ and write it as $$-\frac{1}{4}\cdot\frac{\log(1-2x+2x^2)+\arcsin 2x}{x^3}\cdot\frac{x^2}{\sin^2x}$$ and since the last factor tends to $1$ the desired limit is equal to the limit of the expression $$-\frac{1}{4}\cdot\frac{\log(1-2x+2x^2)+\arcsin 2x}{x^3} $$ Putting $2x=t$ we see that the desired limit equals $$-2\lim_{t\to 0}\frac{\log(1-t+t^2/2)+\arcsin t}{t^3}$$ Now let us observe that applying l'Hospital's Rule will get rid of both $\log $ and $\arcsin $ terms in numerator and will lead us to an algebraic function and that will be easier to handle. On applying l'Hospital's Rule we get the expression -$$-\frac{2}{3}\cdot\frac {(t-1)\sqrt{1-t^2}+1-t+t^2/2}{t^2(1-t+t^2/2)\sqrt{1-t^2}}$$ Since the last two factors in denominator tend to $1$ they can be safely replaced by $1$ and the desired limit equals the limit of expression $$-\frac{2}{3}\cdot\frac{(t-1)\sqrt{1-t^2}+1-t+t^2/2}{t^2}$$ And now we can split using the last term in numerator to get the expression $$-\frac{2}{3}\cdot(1-t)\cdot\frac{1-\sqrt{1-t^2}}{t^2}-\frac{1}{3}$$ The limit of above expression can be easily evaluated (say, by multiplying both numerator and denominator with conjugate) to get $$-\frac {2}{3}\cdot 1\cdot \frac{1}{2}-\frac{1}{3}=-\frac{2}{3}$$

Paramanand Singh
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