How do you compute the limit of #sin(7x)/sin(2x)# as #x->0#?

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Answer 1 · 2016-11-30T15:38:28

#lim_(x->0)frac sin(7x) sin(2x) =7/2#

Considering that:

#lim_(x->0) frac sin(alphax) (alphax) =1#

You can express:

#frac sin(7x) sin(2x) = 7x frac sin(7x) (7x) frac (2x) sin(2x) 1/(2x)#

#frac sin(7x) sin(2x) = 7/2 frac frac sin(7x) (7x) frac sin(2x) (2x) #

and then:

#lim_(x->0)frac sin(7x) sin(2x) = lim_(x->0)7/2 frac frac sin(7x) (7x) frac sin(2x) (2x) =7/2 *1/1 =7/2#