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$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$

Is there a simple way of finding the limit?
I know the long one: rewrite it as $$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$ and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.

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9 Answers 9

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By standard limits

  • $\frac{\sin x}x \to 1$

  • $\frac{1-\cos x}{x^2} \to \frac12$

we have that

$$\frac{\cos x-\cos(3x)}{\sin(3x^2)-\sin(x^2)}=\frac{\frac{\cos x-1+1- \cos(3x)}{x^2}} {\frac{\sin(3x^2)-\sin(x^2)}{x^2}}=\frac{-\frac{1-\cos x}{x^2}+9\frac{1- \cos(3x)}{(3x)^2}} {3\frac{\sin(3x^2)}{3x^2}-\frac{\sin(x^2)}{x^2}}\to\frac{-\frac12+\frac92}{3-1}=2$$

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Hint: Use $$\sin 3a=3\sin a-4\sin^3a$$ $$\cos 3a=4\cos^3a-3\cos a$$ Edit: After substutution it is $$\lim_{x\to0}\frac{2\cos x\sin^2x}{\sin x^2\cos2x^2}=2$$

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    $\begingroup$ Now the votes are balanced! $\endgroup$
    – Nosrati
    Commented Jul 30, 2018 at 21:01
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Hint: Use the factorisation formula $$\cos a -\cos b =-2\sin{a+b\over 2}\sin{a-b\over 2}$$ and $$\sin a -\sin b =2\sin{a-b\over 2}\cos{a+b\over 2}$$

$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \lim_{x\to0}\frac{2\color{red}{\sin 2x}\cdot \color{green}{\sin x}\cdot\color{blue}{x^2}}{\color{blue}{\sin (x^2)}\cos (2x^2)\cdot\color{red}{2x}\cdot \color{green}{x}} =2$$

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I would use Taylor polynomials at order $2$.

$$\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)}=\frac{1-\frac{x^2}{2}-\left(1-\frac{(3x)^2}{2}\right)+o(x^2)}{3x^2-x^2+o(x^2)}=\frac{4x^2+o(x^2)}{2x^2+o(x^2)}=2+o(1) $$

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    $\begingroup$ Minor nitpick: if you're going to use $o$-notation, you should do it throughout the whole calculation, and not only at the end. Otherwise the notation would be inconsistent. $\endgroup$ Commented Jul 30, 2018 at 20:55
  • $\begingroup$ @gimusi Indeed I made an error : not dividing the o() by $x^2$. Thanks for noting. I corrected. $\endgroup$ Commented Jul 31, 2018 at 10:41
  • $\begingroup$ @mathcounterexamples.net Well done! You are welcome, Bye. $\endgroup$
    – user
    Commented Jul 31, 2018 at 13:45
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Simply use Taylor' formula ultimately at order $2$ to find equivalents: near $0$, $$\cos u=1-\frac{u^2}2+o(u^2),\qquad \sin u=u+o(u)$$ so \begin{align} \cos x-\cos 3x&=1-\frac{x^2}2+o(x^2)-\Bigl(1-\frac{9x^2}2+o(x^2)\Bigr)= 4x^2+o(x^2)\\ \sin 3x^2-\sin x^2&=3x^2+o(x^2)-\bigl(\sin x^2+o(x^2)\bigr)=2x^2+o(x^2). \end{align} Thus the numerator is equivalent to $4x^2$, the denominator to $2x^2$, whence $$\frac{\cos x-\cos 3x}{\sin 3x^2-\sin x^2}\sim_0\frac{4x^2}{2x^2}=2.$$

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Use the Maclaurin series $$\cos x = 1-1/2 \; x^2+1/24 \; x^4 +O(x^6)$$ $$\cos 3x = 1-9/2 \; x^2+27/8 \; x^4 +O(x^6)$$ $$\sin x^2 = x^2-1/6\; x^6+O(x^8)$$ $$\sin 3x^2 = 3x^2-9/2 \; x^6+O(x^8)$$ then quotient is $$\frac{-1/2 + 9/2}{3-1} + O(x^2)$$ and therefore the limit is 2.

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  • $\begingroup$ To simplify, It suffices to expand to the second order terms $x^2$. $\endgroup$
    – user
    Commented Jul 30, 2018 at 20:38
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Using l'Hôpital directly is very awkward.

The denominator has $x^2$, but it is immediate that $$ \lim_{t\to0}\frac{\sin3t-\sin t}{t}=3-1=2 $$ Consequently, also $$ \lim_{x\to0}\frac{\sin(3x^2)-\sin(x^2)}{2x^2}=1 $$ Good! Now we can write our limit in the form $$ \lim_{x\to0}\frac{\cos x-\cos 3x}{2x^2}\frac{2x^2}{\sin(3x^2)-\sin(x^2)} $$ The first fraction can be easily dealt with: $$ \lim_{x\to0}\frac{\cos x-\cos 3x}{2x^2}= \lim_{x\to0}\frac{3\sin3x-\sin x}{4x}= \lim_{x\to0}\frac{9\cos3x-\cos x}{4}=2 $$


For completeness, a much simpler strategy is using Taylor expansion: $$ \frac{\cos x-\cos3x}{\sin(3x^2)-\sin(x^2)}= \frac{1-x^2/2-1+(3x)^2/2+o(x^2)}{3x^2-x^2+o(x^2)}= \frac{4+o(1)}{2+o(1)} $$ so the limit is $2$.

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  • $\begingroup$ Is this correct $o(1)→0$? $\endgroup$ Commented Jul 31, 2018 at 0:55
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    $\begingroup$ @TakahiroWaki Yes of course, recall that by definition $o(1)=1\cdot \omega(x)$ with $\omega(x)\to 0$. $\endgroup$
    – user
    Commented Jul 31, 2018 at 5:39
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$$ \cos x-\cos(3x) = \cos x\left(\sin^2x+3\sin^2 x\right) $$

then

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

hence

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

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Start with dividing both the denominator and nominator by $x^2$ and applying quotitent rule for limits, you'll get $$\lim_{x\to 0}\Biggl(\frac{\frac{\cos x-\cos 3x}{x^2}}{\frac{\sin 3x^2-\sin x^2}{x^2}}\Biggl)\space=\space\frac{\lim_{x\to 0}\frac{\cos x}{x^2}-\lim_{x\to 0}\frac{\cos 3x}{x^2}}{\lim_{x\to 0}\frac{\sin 3x^2}{x^2}-\lim_{x\to 0}\frac{\sin x^2}{x^2}}$$ Now remember the particular limits $$\lim_{x\to 0}\frac{\sin x}{x}=1$$ $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$ Then we can manipulate the first expression $$\Biggl(\lim_{x\to 0}\frac{\cos x -1}{x^2}\space+\space\lim_{x\to 0}\frac{1-\cos 3x}{x^2}\Biggl)\space:\space\Biggl(\lim_{x\to 0}\frac{\sin 3x^2}{x^2}\space-\lim_{x\to 0}\frac{\sin x^2}{x^2}\Biggl)\space=\space\Biggl(-\frac{1}{2}\space+\space\frac{9}{2}\Biggl)\space:\space(3-1)\space=\space\frac{8}{2}\space:\space2\space=\space2 $$

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