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

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90
Apr 21, 2017

Answer:

In the Explanation

Explanation:

okay, we have

#lim_(x->0)##(sin2x)/x#

now, if we were to operate it directly, we would get #0/0# as answer which is indefinite, so we need to operate it more before applying the limiting value

So,

#lim_(x->0)##(sin2x)/x#

#= lim_(x->0)##(2sinxcosx)/x# (because #sin2x =2sinxcosx#)

rearrange it as:

#= lim_(x->0)##sinx/x(2cosx)#

#= 1xx2cos (0)# (since#lim_(x->0)##sinx/x=1#)

#=2xx1# (as #cos0=1#)

#=2#

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Write your answer here...
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Answer

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Answer:

Explanation

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32
Oct 20, 2015

Answer:

#2#
Remember : #lim_(x-> 0)(sin kx)/x = k#

Explanation:

PROOF:

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

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

As #x -> 0# : #=> kx-> 0#

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

#=> k * 1 = k#

So here in this case #k =2#

Therefore answer : #2#

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