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Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $

This is what I've tried yet:

$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\ =&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\ =& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\ =& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\ =& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\ =& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$

Taking limit of the above expression, we've :

$$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$

Firstly, is my answer right or am I doing somewhere wrong? Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred?

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  • $\begingroup$ Well, are you allowing L'Hôpital's Rule or power series? Either of those is going to make this shorter. $\endgroup$ – Chappers May 3 '15 at 2:06
  • $\begingroup$ Yep! There are no limits mentioned, neither in the book nor in my question to get it solved. Any way shall work. I'll try to do it with L'H rule.. .can you show me how to do it with power series? $\endgroup$ – Kushashwa Ravi Shrimali May 3 '15 at 2:08
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    $\begingroup$ After all of the great answer that used series expansions, I thought I would take a different tact. Let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola May 3 '15 at 5:18
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Several answers used Taylor expansions to establish the limit.

Just as a long comment, let me show that, using one or two extra terms in the developments, you can get more information. $$A= \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$$ and let use $$\tan(y)=y+\frac{y^3}{3}+\frac{2 y^5}{15}+\frac{17 y^7}{315}+O\left(y^9\right)$$ So, the numerator is $$x(2 x+\frac{8 x^3}{3}+\frac{64 x^5}{15}+\frac{2176 x^7}{315}+O\left(x^9\right))-2x(x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right))$$ that is to say $$2 x^4+4 x^6+\frac{34 x^8}{5}+O\left(x^{10}\right)$$ Similarly, for the denominator, we have $$4 x^4-\frac{8 x^6}{3}+\frac{4 x^8}{5}+O\left(x^9\right)$$ So, for the ratio $$A=\frac{2 x^4+4 x^6+\frac{34 x^8}{5}+O\left(x^{10}\right)}{4 x^4-\frac{8 x^6}{3}+\frac{4 x^8}{5}+O\left(x^9\right)}$$ Performing the long division, we then get $$A=\frac{1}{2}+\frac{4 x^2}{3}+\frac{112 x^4}{45}+O\left(x^5\right)$$ So, not only you get the limit but you also see how it is approached.

I you plot on the same graph the function and the above approximation, you will probably be amazed to notice how close to each other are the two curves for $-\frac 12 \leq x \leq \frac 12$.

Now, suppose that you need an approximate solution of the equation $$\cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}=1$$ Using the approximation, you then need to solve the quadratic (in $x^2$) $$\frac{1}{2}+\frac{4 x^2}{3}+\frac{112 x^4}{45}=1$$ the roots of which being $$x_{\pm}=\pm \frac{1}{2} \sqrt{\frac{3}{14} \left(\sqrt{95}-5\right)}\approx \pm 0.504274$$ while the exact solutions are $\pm 0.473874$.

Edit

Since you looked to be interested by this approach, let me tell you (for your curiosity) than we can do much better using, insteat of Taylor expansions, Pade approximants (you will learn about them rather soon - they are ratio of polynomials). Applied to the problem you posted, we should obtain $$A \approx \frac{\frac{40 x^4}{693}+\frac{26 x^2}{99}+\frac{1}{2}}{\frac{8816 x^4}{10395}-\frac{212 x^2}{99}+1}$$ Performing the long division, you should get the same approximation as before. But, plot this function with the other two; this last one almost coincide with the original function.

If we use this last approximation for solving the equation $A=1$, the roots will be $$x_{\pm}=\pm\frac{1}{2} \sqrt{\frac{3 \left(4165-\sqrt{12602485}\right)}{2054}}\approx \pm 0.473880$$ very close to the exact answer.

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    $\begingroup$ Indeed! I'm really amazed. Loved how you spared time to go deeper into the solution! Hats off to your spirit, Mr. Claude! Thanks a ton! :) $\endgroup$ – Kushashwa Ravi Shrimali May 3 '15 at 3:30
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    $\begingroup$ You are very welcome ! But, I must (once more) confess that I flet in love with Taylor series 60 years ago. $\endgroup$ – Claude Leibovici May 3 '15 at 3:34
  • $\begingroup$ Nice answer, as usual! I wonder though how you computed the Pade approximants? Not by hand right? $\endgroup$ – user21820 May 3 '15 at 9:55
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Well, let's try something different from using power series expansions. Here, we simplify using trigonometric identities to reveal that

$$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{2x}{\sin 4x }=\frac{1}{2\text{sinc}(4x)}$$

where the sinc function is defined as $\text{sinc}(x)=\frac{\sin x}{x}$.

The limit as $x \to 0$ is trivial since $\text{sinc}(4x) \to 1$ . The limit is $1/2$ as expected.


NOTE $1$: Establishing the identity

Using standard trigonometric identities, we can write

$$\begin{align} x\tan 2x-2x \tan x&=\frac{2x\sin x\cos x}{\cos 2 x}-2x\frac{\sin x}{\cos x}\\\\ &=2x \sin x \frac{\sin^2x}{\cos x\cos 2x}\\\\ &=2 \frac{\sin^4 x}{\text{sinc}( 4 x)} \end{align}$$

and

$$(1-\cos 2x)^2=4\sin^4 x$$

Putting it together reveals that

$$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{1}{2\text{sinc}(4x)}$$


NOTE $2$: Series expansion is facilitated by simplifying using trigonometric identities

We can use the Laurent series for the cosecant function

$$\csc x=\frac1 x+\frac16 x+\frac{7}{360}x^3+\frac{31}{15120}x^5+O(x^7)$$

to establish that

$$\begin{align} \frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}&=\frac{2x}{\sin 4x }=2x\text{csc}(4x)\\\\ &=\frac12 +\frac43 x^2 +\frac{112}{45}x^4+\frac{3968}{945}x^6+O(x^7) \end{align}$$

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Ok this is a better way of doing this:

$$\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $$

$$= \lim_{x\to 0} \frac{1}{4} \frac{x}{\sin x} \frac{\tan 2x - 2\tan x}{\sin^3 x}$$

$$= \lim_{x\to 0} \frac{1}{4} \frac{\sin2x \cos x - 2\sin x \cos 2x}{\cos 2x \cos x \sin^3 x}$$

$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos 2x}{\cos 2x \cos x \sin^ 2 x}$$

$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos^2 x + \sin^2x}{\cos2x \cos x \sin^2 x}$$

$$ = \lim_{x\to 0} \frac{1}{2} \frac{1}{\cos 2x \cos x} = \frac{1}{2}$$

So your answer was right, but I feel like this is a faster and neater way of doing this.

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  • $\begingroup$ That looks better than my solution. Thanks a lot for helping me out! $\endgroup$ – Kushashwa Ravi Shrimali May 3 '15 at 3:32
  • $\begingroup$ @KushashwaRaviShrimali NP. $\endgroup$ – SalmonKiller May 3 '15 at 3:33
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Using power series, we only need two: $$ \tan{x} = x + \frac{1}{3}x^3 + O(x^5), $$ and $$ \cos{x} = 1- \frac{1}{2}x^2 + O(x^4). $$ Then $$ x(\tan{2x}-2\tan{x}) = x\left( 2x+\frac{8}{3}x^3 - 2x - \frac{2}{3}x^3 + O(x^5) \right) = 2 x^4 + O(x^5), $$ and $$ (1-\cos{2x})^2 = \frac{1}{4}(2x)^4 + O(x^6) = 4x^4 + O(x^6), $$ and then just divide to find the limit as $1/2$.

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  • $\begingroup$ Great! Thanks a lot for your help, Chappers! $\endgroup$ – Kushashwa Ravi Shrimali May 3 '15 at 2:22
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Your solution looks good! Give yourself a pat on your back.

Another way is to expand the trigonometric polynomials using Taylor series, i.e., \begin{align} \dfrac{x(\tan(2x)-2\tan(x))}{(1-\cos(2x))^2} & = x \cdot \dfrac{2x+ \dfrac{(2x)^3}3+ \mathcal{O}(x^5) - 2x - 2\cdot\dfrac{x^3}3 + \mathcal{O}(x^5)}{\left(1-\left(1-\dfrac{(2x)^2}{2!} + \mathcal{O}(x^4)\right)\right)^2}\\ & = x \cdot \dfrac{2x^3 + \mathcal{O}(x^5)}{\left(2x^2 + \mathcal{O}(x^4)\right)^2} = \dfrac{2x^4\left(1+\mathcal{O}(x^2)\right)}{4x^4\left(1+\mathcal{O}(x^2)\right)^2} \end{align} Taking the limit as $x \to 0$, you get the limit as $\dfrac12$.

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  • $\begingroup$ Thanks for the appreciation. And thanks a lot for your kind help. $\endgroup$ – Kushashwa Ravi Shrimali May 3 '15 at 2:23

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