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how can u find the limit for (sin2x)/x as x approaches 0?
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L'Hôpital's rule
don't know what that is im in trig
i learned it by finding the sin of 2x in degrees first
\[\lim_{x \rightarrow 0} \frac{\sin(2x)}{x}\] Differentiate numerator and denominator \[ \lim_{x \rightarrow 0} \frac{2\cos(2x)}{1}\] As x approaches zero, \[= \lim_{x \rightarrow 0} 2\cos(0) \\ \\ =\lim_{x \rightarrow 0} 2(1) \\ \\ =2\]
where did you get cosine from?
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You differentiate sin(2x) you'll get 2cos(2x)
Without using L Hopital's Rule, you want to use the fact that:\[\lim_{u\rightarrow 0}\frac{\sin(u)}{u}=1\] Does this look familiar?
nope i was taught to find the degree value of sin 2x first? idk .....
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