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)
 
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