1

I can get the correct answer through one solution, but when I try the second method, it shows an obvious error, and I can't find where and why. Can someone know the reason, or can provide some useful suggestions? thanks, : )

Solution2

$$\begin{align*} \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}} & = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{({\rm{1 + }}\cos 2x{\rm{ - 1}})}^{\frac{1}{2}}}{{({\rm{1 + }}\cos 3x{\rm{ - 1}})}^{\frac{1}{3}}}}}{{{x^2}}}\\ & =\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x({\rm{1 + }}\frac{{\cos 2x{\rm{ - 1}}}}{{\rm{2}}})({\rm{1 + }}\frac{{\cos 3x{\rm{ - 1}}}}{{\rm{3}}})}}{{{x^2}}}\\ &=\mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x}}{{{x^2}}}\\ &=\frac{{\rm{1}}}{{\rm{2}}} \end{align*}$$

Maybe Solution1 is a bit informal, but all I want about it is just to talk about ideas

Solution1

$$\begin{align*} \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{{{x^2}}} & = \mathop {\lim }\limits_{x\to0} \frac{{1 - \cos x + \cos x(1 - {{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\ &=\frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(\cos 2x)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(\cos 3x)}^{\frac{1}{3}}})}}{{{x^2}}}\\ &= \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - {{(1 + \cos 2x - 1)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(1 + \cos 3x - 1)}^{\frac{1}{3}}})}}{{{x^2}}}\\ & = \frac{1}{2}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 2x}}{{2{x^2}}}\mathop { + \lim }\limits_{x\to0} \frac{{1 - \cos 3x}}{{3{x^2}}}\\ &= \frac{1}{2} + 1 + \frac{3}{2} = 3 \end{align*}$$

I find my mistake is "forget considering the infinitesimal term when replacing"

thanks help for clear answer for @user

a wonderful and general answer for @CHAMSI

also, Parthib Ghosh's opinion is also useful

xioacd99
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3 Answers3

3

When calculating limits, you can not just replace $ x $ in a part of the expression and leave it in the other part, this is a very common mistake.

I'll provide a solution to a more generalized limit, and won't use series expansions or L'hopital's rule. We must know that $ \frac{1-\cos{x}}{x^{2}}\underset{x\to 0}{\longrightarrow}\frac{1}{2} $, though.

Let $ n\in\mathbb{N}^{*}\left(=\mathbb{N}\setminus\left\lbrace 0\right\rbrace\right) $, we have : \begin{aligned}\lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\lim_{x\to 0}{\frac{\sum\limits_{p=1}^{n}{\left(\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}-\prod\limits_{k=1}^{p}{\cos^{1/k}{\left(kx\right)}}\right)}}{x^{2}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos^{1/p}{\left(px\right)}}{x^{2}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\&=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{\left(1-\cos^{1/p}{\left(px\right)}\right)\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}{x^{2}\color{blue}{\times\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\frac{1-\cos{\left(px\right)}}{x^{2}\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\prod_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}}\\ &=\lim_{x\to 0}{\sum_{p=1}^{n}{\left(p^{2}\times\frac{1-\cos{\left(px\right)}}{\left(px\right)^{2}}\times\frac{\prod\limits_{k=1}^{p-1}{\cos^{1/k}{\left(kx\right)}}}{\sum\limits_{j=0}^{p-1}{\cos^{j/p}{\left(px\right)}}}\right)}}\\&=\sum_{p=1}^{n}{\left(p^{2}\times\frac{1}{2}\times\frac{\prod\limits_{k=1}^{p-1}{1}}{\sum\limits_{j=0}^{p-1}{1}}\right)}\\ &=\frac{1}{2}\sum_{p=1}^{n}{p}\\ \lim_{x\to 0}{\frac{1-\prod\limits_{k=1}^{n}{\cos^{1/k}{\left(kx\right)}}}{x^{2}}}&=\frac{n\left(n+1\right)}{4}\end{aligned}

CHAMSI
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1

It should be noted that

$\color{blue}{\left(1+x\right)^{\frac{1}{2}}=1+\dfrac{x}{2}+\cdots}$

is true for $\color{orange}{|x|\le1}$

In your assumption, $\mathtt{\big(1+cos(2x)-1\big)^{\frac{1}{2}}=1+\dfrac{cos(2x)-1}{2}},$

$\mathtt{\color{violet}{|cos(2x)-1|\le2}}$

So, I think it should not be used.

Instead, use binomial expansion for $\cos(x)$

Parthib Ghosh
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  • 1
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  • Note that we are looking at the limit $x\to 0$. If $x$ is close to $0$, $\cos(2x)$ is close to $1$, i.e., $\cos(2x)-1$ is close to $0$, so the binomial expansion is a perfectly fine choice. What do you mean by your last sentence? – Gary Oct 09 '21 at 06:32
  • I mean $\cos(x)=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots$ – Parthib Ghosh Oct 09 '21 at 06:39
  • That is Taylor (or Maclaurin) expansion not binomial. – Gary Oct 09 '21 at 06:40
  • By the way, if you put $$ before and after an equation, it will be displayed, you will not need a displaystyle command. – Gary Oct 09 '21 at 06:51
1

In the solution $1$ at this step

$$ \ldots= \frac{1}{2}\mathop { + \lim }\limits_{x \to 0} \frac{{1 - {{(1 + \cos 2x - 1)}^{\frac{1}{2}}}{\rm{ + }}{{(\cos 2x)}^{\frac{1}{2}}}(1 - {{(1 + \cos 3x - 1)}^{\frac{1}{3}}})}}{{{x^2}}} \ldots$$

it seems you are make the substitution $\cos x =1$ which is not allowed and also in the subsequent step what you did is not much clear.

Solution $2$ is almost fine but using binomial expansion we need also to consider the remainder term, that is for example

$$\lim_{x \to 0} \frac{\left(1+(\cos x-1)\right)^\frac12}{x^2}=\lim_{x \to 0} \frac{\left(1+\frac12(\cos x-1)+O((\cos x-1)^2)\right)}{x^2}$$

Take the limit for single part of the expression is not allowed in general and even if in some case it leads to a correct result you should take caution with these kind of steps.

Refer also to the related

Edit

To take the limit using your idea we can proceed as follows

$$ \frac{{1 - \cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{x^2}=\frac{{1 - \cos x + \cos x-\cos x{{(\cos 2x)}^{\frac{1}{2}}}{{(\cos 3x)}^{\frac{1}{3}}}}}{x^2}=$$

$$=\frac{1-\cos x}{x^2} + \cos x\frac{1-(\cos 2 x)^\frac 12 (\cos 3x)^\frac 13}{x^2} \to \frac12+1\cdot \frac 5 2=3$$

indeed

$$\frac{1-(\cos 2 x)^\frac 12 (\cos 3x)^\frac 13}{x^2}=\frac{1-(1+(\cos 2 x-1))^\frac 12 (1+(\cos 3x-1))^\frac 13}{x^2}=$$

$$=\frac{1-\left(1+\frac{\cos 2 x-1}2+O((\cos 2x-1)^2)\right) \left(1+\frac{\cos 3 x-1}3+O((\cos 3x-1)^2)\right)}{x^2}=$$

$$=\frac{\frac{1-\cos 2 x}2+\frac{1-\cos 3 x}3-\frac{\cos 2 x-1}2\frac{\cos 3 x-1}3+O(x^4)}{x^2}=$$

$$=\frac{3(1-\cos 2 x)+2(1-\cos 3 x)-(\cos 2x -1) (\cos 3x-1)+O(x^4)}{6x^2}=$$

$$=\frac{2-2\cos 2x +1-\cos 3x +1-\cos 2x \cos 3x+O(x^4)}{6x^2}=$$

$$=\frac43 \frac{1-\cos 2x}{4x^2}+\frac32\frac{1-\cos 3x}{9x^2}+\frac{1-\cos 2x\cos 3x}{6x^2}+O(x^2) \to \frac23+\frac 34+\frac{13}{12}=\frac52 $$

indeed

$$\frac{1-\cos 2x\cos 3x}{6x^2}=\frac{(1-\cos 2x)(\cos 3x-1)+2-\cos 2x-\cos 3x}{6x^2}=$$

$$=\frac{1-\cos 2x}{2x}\frac{\cos 3x-1}{3x}+\frac23\frac{1-\cos 2x}{4x^2}+\frac32\frac{1-\cos 3x}{9x^2}\to 0+\frac13+\frac34 =\frac{13}{12}$$

Edit 2

The limit can be solved in a more effective way using that

  • $\frac{1-\cos x}{x^2} \to \frac12 \iff \cos x= 1-\frac12 x^2 +o(x^2)$

therefore

  • $(\cos 2x)^{\frac12}=\left(1-\frac12 (2x)^2 +o(x^2)\right)^\frac12=1- x^2 +o(x^2)$
  • $(\cos 3x)^{\frac13}=\left(1-\frac12 (3x)^2 +o(x^2)\right)^\frac13=1- \frac 32 x^2 +o(x^2)$

and

$$1 - \cos x (\cos 2x)^{\frac12}(\cos 3x)^{\frac13}=$$ $$=1-\left(1-\frac12 x^2 +o(x^2)\right)\left(1- x^2 +o(x^2)\right)\left(1-\frac32 x^2 +o(x^2)\right)=$$

$$=\frac12 x^2+x^2+\frac32 x^2 +o(x^2)=3x^2+o(x^2)$$

which leads to

$$ \frac{ 1 - \cos x (\cos 2x)^{\frac12}(\cos 3x)^{\frac13}}{x^2}=\frac{3x^2+o(x^2)}{x^2}=3+o(1) \to 3$$

user
  • 154,566
  • About this step

    $\begin{array}{l} 1 - cosx{(cos2x)^{\frac{1}{2}}}{(cos3x)^{\frac{1}{3}}} = \ = 1 - (1 - \frac{1}{2}{x^2} + o({x^2}))(1 - {x^2} + o({x^2}))(1 - \frac{3}{2}{x^2} + o({x^2})) = \ = \frac{1}{2}{x^2} + {x^2} + \frac{3}{2}{x^2} + o({x^2}) = 3{x^2} + o({x^2}) \end{array}$

    Are there any formulas or properties used here?

    I multiply them one by one, and the result is as follows

    $ - \frac{{4{x^6}{o^3} + 11{x^6}o - 12{x^6}{o^2} - 3{x^6} + 12{x^4}{o^2} + 11{x^4} - 24{x^4}o + 12{x^2}o - 12{x^2}}}{4}$

     – xioacd99 Oct 10 '21 at 01:49
  • @xioacd99 The formula (i.e. little-o property) is that $x^no^m(x^2)=o(x^2)$ and $o(x^2)+ o(x^2)= o(x^2)$. Therefore we don’t need to expand all the terms, we can put all higher order terms into one $ o(x^2) $ term. – user Oct 10 '21 at 05:32
  • Take a look here and here. – user Oct 10 '21 at 05:38
  • @xioacd99 Indeed in the full expression you derived dividing by $x^2$ and taking the limit we obtain $3$ because by definition $\frac{o(x^2)}{x^2} \to 0$. – user Oct 10 '21 at 06:27
  • put all higher order terms into one $o(x^2)$ is wonderful, I was stuck in the traditional way of thinking before, thanks – xioacd99 Oct 10 '21 at 08:21
  • Yes it is indeed a very useful notation! Bye – user Oct 10 '21 at 09:45