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Bomull
lim(x->0) sin(x)/sin(2x) ?
\[\lim (x->0) \frac{ \sin(x) }{ \sin(2x) }\] ?
\[\lim_{x\to 0}\frac{\sin x }{\sin 2x}\] We know that \[\lim_{x\to 0}\frac{\sin x }{x}=1\] Could you use this here to solve ? @Bomull
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?
\[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{x}{\sin 2x}\] 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 \[\lim_{x\to 0} \frac{\sin x}{x}\times \frac{2x}{\sin 2x}\times \frac 1 2\] \[1\times 1 \times \frac 1 2=\frac 1 2\]
Do you understand this?
ah ok thanks! \[\lim_{x \rightarrow 0} \frac{ nx }{ \sin nx }\] is good to know. I think I got it