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$$ \lim_{x\to 0}\dfrac{\sin\left(\sin\left(2x\right)\right) - 2x} {\tan\left(2x\right) - \sin\left(2x\right)} $$ This problem is from a past exam in our calculus course.

  • Since we weren’t taught the Maclaurin expansion for $\tan\left(u\right)$, I had to solve it by multiplying both the numerator and denominator by $\cos\left(2x\right)$ and using the expansions for $\sin\left(u\right)$ and $\cos\left(u\right)$ that we were taught: It was quite cumbersome.

I wondered if there might be a simpler way that avoids the expansion for $\tan\left(u\right)$. Thank you !.

Sine of the Time
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ten_to_tenth
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2 Answers2

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Hint:

$$\dfrac{\sin(\sin2x)-2x}{\tan2x-\sin2x}$$ $$=\cos2x\cdot\dfrac{\dfrac{\sin(\sin2x)-\sin2x}{x^3}-\dfrac{\sin2x-2x}{x^3}}{\dfrac{\sin2x(1-\cos2x)}{x^3}}$$ $$=\cos2x\cdot\dfrac{\left(\dfrac{\sin2x}x\right)^3\cdot\dfrac{\sin(\sin2x)-(\sin2x)}{(\sin2x)^3}-8\cdot\dfrac{\sin2x-2x}{(2x)^3}}{\dfrac{\sin2x(2\sin^2x)}{x^3}}$$

Then utilize: Are all limits solvable without L'Hôpital Rule or Series Expansion to find $$\lim_{y\to0}\dfrac{y-\sin y}{y^3}=\dfrac16$$

lab bhattacharjee
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  • Thank you very much! I learned from your solution that identifying the lowest-order term in the difference helps determine what to divide the numerator and denominator by. After that, the solution becomes much more straightforward. – ten_to_tenth Jan 15 '25 at 05:13
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We can use that

$$\frac{\sin\left(\sin\left(2x\right)\right) - 2x} {\tan\left(2x\right) - \sin\left(2x\right)}=\frac{\cos\left(2x\right)}{\sin\left(2x\right)}\;\frac{\sin\left(\sin\left(2x\right)\right) - 2x} {1 - \cos\left(2x\right)}=$$

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

using standard limits and for the latter factor

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

user
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