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Bomull Group TitleBest ResponseYou've already chosen the best response.0
\[\lim (x>0) \frac{ \sin(x) }{ \sin(2x) }\] ?
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{x\to 0}\frac{\sin x }{\sin 2x}\] We know that \[\lim_{x\to 0}\frac{\sin x }{x}=1\] Could you use this here to solve ? @Bomull
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.0
I don't know how..
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Okay, I'll explain you \[\lim_{x\to 0} \frac{\sin x}{\sin 2x}\] Let's multiply and divide by x \[\lim_{x\to 0} \frac{\sin x}{\sin 2x}\times \frac x x\] \[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}\] Do you understand till here?
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
\[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}\] Let's multiply and divide by 2 \[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2\] We know that \[\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{nx}{\sin nx}=1\] so we get \[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2\] \[1\times 1 \times \frac 1 2=\frac 1 2\]
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Do you understand this?
 one year ago

Bomull Group TitleBest ResponseYou've already chosen the best response.0
ah ok thanks! \[\lim_{x \rightarrow 0} \frac{ nx }{ \sin nx }\] is good to know. I think I got it
 one year ago
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