Find the limit as x-0
(sin 2x)/(4x)

Question

Find the limit as x->0
(sin 2x)/(4x)

anonymous · Answer

1/2?

psymon · Answer

Answer given to ya, so no real point.

anonymous · Answer

how is that?

anonymous · Answer

how is it given? .--. lol

psymon · Answer

As in the answer was just posted, so seemed like there was no explanation needed anymore.

anonymous · Answer

no no, please explain! lol I need to understand how this is worked out! :p

psymon · Answer

Alright. So you know the identity:

$$\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1$$ correct?

anonymous · Answer

yes...

psymon · Answer

Alrighty, cool. So, that x can be anything, as long as its not an x in the denominator of a fraction. So I could have sin(lnx)/lnx, still 1. sin(e^(pi))/e^(pi), still = 1, as long as the angle of sin and the denominator underneath it match. So as of now, you have sin2x, meaning we want whatever is underneath it to also be 2x. But right now its stuck as 4x. So basically, we need to multiply by something in order to turn 4x into 2x.

anonymous · Answer

oh okay! so you multiply 1/2?..

psymon · Answer

Basically. But you have to do it by the top and bottom:

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