[ASK] Limit of Trigonometry Function

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SUMMARY

The limit of the trigonometric function $$\lim_{x\to0}\frac{sin2x+sin6x+sin10x-sin18x}{3sinx-sin3x}$$ evaluates to 192. The discussion highlights that both substitution and L'Hôpital's Rule initially yield an indeterminate form of $$\frac{0}{0}$$. By applying L'Hôpital's Rule multiple times, specifically until the third derivative, the limit simplifies to $$\frac{4,608}{24}$$, confirming the result as 192.

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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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