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Determine the limit of the trig . function. What should I do first ?
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\[\lim_{x \rightarrow 0} \frac{ (\sin 2x) }{ \sin3x }\]
u know the lim x=> 0 sin x/x =1 ??
yes, what do i do with the 2x&3x ?
first divide numnerator and denominator by x, and tell me what u get..
(sin 2)/(sin3x)
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lol no! like this \(\huge \frac{\frac{sin 2x}{x}}{\frac{sin3x}{x}}\) got this?
so, \[\frac{ 2(\sin2x/2x) }{ 3(\sin3x/3x) }\] then \[\frac{ 2(1) }{ 3(1) }\]?
absolutely !
thanks!
welcome ^_^
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