# How do you find the limit of #(sin2x)/x# as x approaches 0?

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Apr 21, 2017

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In the Explanation

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okay, we have

now, if we were to operate it directly, we would get

So,

rearrange it as:

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Oct 20, 2015

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Remember :

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PROOF:

#lim_(x-> 0)(sin kx)/x = lim_(x-> 0) (k * (sin kx)/(kx))#

#= k * lim_(x-> 0)(sin kx )/ (kx)#

As

#=> k * lim_(kx -> 0) (sin kx)/(kx)#

#=> k * 1 = k#

So here in this case

Therefore answer :

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