MHB [ASK] Limit of Trigonometry Function

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The limit of the trigonometric function given is evaluated as $$\lim_{x\to0}\frac{sin2x+sin6x+sin10x-sin18x}{3sinx-sin3x}$$. Initial attempts using substitution or L'Hôpital's rule yield an indeterminate form of $$\frac{0}{0}$$. Further simplification leads to the third derivative, resulting in a final answer of 192. The calculation confirms that the limit can be resolved through repeated application of L'Hôpital's rule. Thus, the correct answer is 192.
Monoxdifly
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$$\lim_{x\to0}\frac{sin2x+sin6x+sin10x-sin18x}{3sinx-sin3x=}$$
A. 0
B. 45
C. 54
D. 192
E. 212

Either substituting or using L'Hopital gives $$\frac00$$. Is there any way to simplify it and make the result a real number?
 
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We can repeat L'Hôpital until it's not $\frac 0 0$ any more.
 
Ah, I see. It's in the third derivative and the answer is $$\frac{4,608}{24}$$= 192, right?
 
Monoxdifly said:
Ah, I see. It's in the third derivative and the answer is $$\frac{4,608}{24}$$= 192, right?
Yep. (Nod)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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