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I'm a bit confused regarding this question. I've been trying to solve it and have gotten to the same answer ($3/8$) thrice now. I have no idea where I'm going wrong and would really appreciate some help figuring it out. Here's my solution:

$$\lim_{x \to 0}\frac{x\tan2x-2x\tan x}{(1-\cos2x)^2}$$ Applying L'Hospital's rule,

$$\lim_{x \to 0}\frac{\tan2x+x(\sec^22x)(2)-2\tan x-2x\sec^2x}{2(1-\cos2x)(\sin2x)(2)}$$ Dividing numerator and denominator by $2x$, $$\begin{align*} \lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\ &=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\ &=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)} \end{align*}$$ Applying L'Hôpital's rule (again), $$\lim_{x \to 0}\frac{(2)(\sec2x)(\sec2x)(\tan2x)(2)-(2)(\sec x)(\sec x\tan x)}{4(\sin2x)(2)}$$ $$=\lim_{x \to 0}\frac{(2)(\sec^22x)(\tan2x)-(\sec^2x)(\tan x)}{4(\sin2x)}$$ Dividing numerator and denominator by $2x$, $$\begin{align*} &=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{2x})}{4(\frac{\sin2x}{2x})}\\ &=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{x})(\frac{1}{2})}{4(\frac{\sin2x}{2x})}\\ &=\lim_{x \to 0}\frac{(2)(\sec^22x)(1)-(\sec^2x)(1)(\frac{1}{2})}{4(1)}\\ &=\frac{(2)(\sec^20)(1)-(\sec^20)(1)(\frac{1}{2})}{4(1)}\\ &=\frac{(2)(1)(1)-(1)(1)(\frac{1}{2})}{4(1)}\\ &=\frac{2-\frac{1}{2}}{4}\\ &=\frac{\frac{3}{2}}{4}\\ &=\frac{3}{8}. \end{align*}$$

Bernard
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somecoolusername
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  • Please do not make subject lines consist entirely of mathjax, as it messes up navigation shortucts. – Arturo Magidin Nov 23 '21 at 19:03
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    im sorry, i have no idea about mathjax whatsoever and only copied a few syntaxes to write the question. I'll try to avoid making the same mistake again. thanks for your input! :D – somecoolusername Nov 23 '21 at 19:11
  • The numerator is $\propto 2x-2x^2$ and the denominator $\propto x^4$. –  Nov 23 '21 at 19:12
  • The title was changed since. –  Nov 23 '21 at 19:17
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    Remark: Transforming your denominator to $4\sin^4(x)$ and using the fact $\sin(x)/x\to 1$ as $x\to 0$ enables one to rewrite your limit as $\frac14\cdot\lim_{x\to 0}\frac{\tan(2x)-2\tan(x)}{x^3}$ which is much easier to evaluate. – Marian G. Nov 23 '21 at 19:38
  • You can't replace $\frac{\tan kx}{kx}$ with $1$ before you differentiate again. The correct answer is $\frac12$. – J.G. Nov 23 '21 at 19:42
  • @J.G. why not? i mean, $lim_{x \to 0}(\frac{\tan kx}{kx})$ is $1$, isn't it? Also, $lim_{x \to 0} f(x)-g(x)= lim_{x \to 0}f(x)-lim_{x \to 0}g(x)$ so where am i going wrong? – somecoolusername Nov 24 '21 at 06:07
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    The mistake is made in the 3rd step by taking the limit partially on individual terms. For example: $$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^{3}} \quad\left(\frac{0}{0}\right)\ =& \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^{2}} \ =& \frac{1}{3} \lim _{x \rightarrow 0} \frac{\sin x}{x} \ =& \frac{1}{3} \end{aligned} \ \text{On the other hand, }$$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\frac{\sin x}{x}-\cos x}{x^{2}} \ =& \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}} \ =& \frac{1}{2} \ \neq & \frac{1}{3} \end{aligned} $$ – Lai Nov 24 '21 at 09:04

5 Answers5

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Your problem arises here: \begin{align*} \lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\ &=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\ &=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)} \end{align*}

particularly where you evaluate $\lim_\limits{x\to 0} \frac {\tan 2x}{2x} = 1$ and $\lim_\limits{x\to 0} -\frac {2\tan x}{2x} = -1$ and cancel them.

$\tan 2x - 2\tan x\approx 2x^3$ and that is not trivial.

user317176
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  • Yeah, but $\lim_\limits{x\to 0} \frac {\tan 2x}{2x} - \lim_\limits{x\to 0} \frac {2\tan x}{2x} \approx \lim_\limits{x\to 0} \frac {2x^3}{2x} = \lim_\limits{x\to 0} x^2= 0$ – somecoolusername Nov 24 '21 at 06:15
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    Except, what you have is: $\frac {\frac {\tan 2x}{2x} - \frac {2\tan x}{2x}}{4(1-\cos 2x)} = \frac {x^2+ O(x^4)}{8x^2 + O(x^4)}.$ In general, when you have $\lim_\limits {x\to a} f(x) = \lim_\limits {x\to a} g(x)$ and $\lim_\limits {x\to a} h(0)$ you cannot say $\lim_\limits {x\to a} \frac {f(x) - g(x)}{h(x)} = 0$ – user317176 Nov 24 '21 at 17:44
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It's a small conceptual mistake after applying L'Hospital's rule for the 1st time then "dividing Nr and Dr by 2x and then applying the limits individually without expressing limit as a sum."

Do note that after dividing the Nr and Dr with 2x and then applying the limits individually you get a completely new function not same as the one before i.e. line 2 and line 3 are two different function.

You can still solve this without much hard work.. just express everything in terms of tanx..

Ayush
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  • How come? I mean, i can still cancel out 2x from both Nr and Dr and get the same function, though... – somecoolusername Nov 24 '21 at 06:10
  • You can verify that when you plot both the functions.. they are two different functions.. – Ayush Nov 24 '21 at 06:15
  • i plotted them and they seem to be completely identical. You're talking about the functions before and after diving Nr and Dr by $2x$ for the first time, right? – somecoolusername Nov 24 '21 at 06:26
  • Also a very common mistake here is.. when you use the result lim x->0 f(x)/x =a(constant or any fixed result) after expressing the limit as a sum and then as product of two limits .. make sure that the other terms are in indeterminant form – Ayush Nov 24 '21 at 06:38
  • https://www.desmos.com/calculator/gfzp2lixw0 You can check verify that here what I was talking about – Ayush Nov 24 '21 at 06:46
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Hint:

Another way:

Observe that

$$x\tan2x-2x\tan x=x(\tan2x-2x)-2x(\tan x-x)$$

Use Are all limits solvable without L'Hôpital Rule or Series Expansion to find $$\lim_{h\to0}\dfrac{\tan h-h}{h^3}=\dfrac13$$

Can you take it from here?

lab bhattacharjee
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I am now going to find the limit without L’Hospital Rule. $$\begin{array}{l} \displaystyle \quad \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}\\ =\displaystyle \lim _{x \rightarrow 0} \frac{\frac{x \sin 2 x}{\cos 2 x}-\frac{2 x \sin x}{\cos x}}{(1-\cos 2 x)^{2}}\\ =\displaystyle \lim _{x \rightarrow 0} \frac{x \sin 2 x \cos x-2 x \sin x \cos 2 x}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\ =\displaystyle \lim _{x \rightarrow 0} \frac{x\left(\sin 2 x \cos x-\sin x \cos 2 x\right)-x \sin x \cos 2 x}{\tan 2 x \cos x(1-\cos 2 x)^{2}}\\ =\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x(1-\cos 2 x)}{\cos 2 x \cos x(1-\cos 2 x)^{2}}\\ =\displaystyle \left(\lim _{x \rightarrow 0} \frac{x \sin x}{1-\cos 2 x}\right)\left(\lim _{x \rightarrow 0} \frac{1}{\cos 2 x \cos x}\right)\\ =\displaystyle \lim _{x \rightarrow 0} \frac{x \sin x}{2 \sin ^{2} x}\\ =\displaystyle \frac{1}{2} \lim _{x \rightarrow 0} \frac{x}{\sin x}\\ =\displaystyle \frac{1}{2} \end{array}$$

Lai
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  • mhmm! i did manage to get to $\frac{1}{2}$ as well using a different approach. I just couldn't figure out why the solution i mentioned was wrong (still can't :/). Thanks for your input, though!! :D – somecoolusername Nov 24 '21 at 06:18
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    Please read my comment above in your question. – Lai Nov 24 '21 at 09:15
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Hint:

Use Taylor-Young's formula at (the final) order $4$ and do some trigonometry:

  • $\tan x= x+\dfrac{x^3}3+o(x^3)$;
  • $(1-\cos2x)^2=(2\sin^2 x)^2=4\sin^4x=4\bigl(x+o(x)\bigr)^4$.
Bernard
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