How do you find the limit of # ( x+sin 2x)/( 3x) # as x approaches 0?

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Answer 1 · 2016-07-01T15:58:02

#Lt_(x->0)(x+sin2x)/(3x)=1#

#Lt_(x->0)(x+sin2x)/(3x)#

= #Lt_(x->0)(x/(3x)+(sin2x)/(3x))#

= #Lt_(x->0)(1/3+(sin2x)/(2x)xx(2x)/(3x))#

= #Lt_(x->0)(1/3)+Lt_(x->0)(sin2x)/(2x)xxLt_(x->0)(2x)/(3x)#

= #1/3+Lt_(2x->0)(sin2x)/(2x)xx2/3#

= #1/3+1xx2/3#

= #1/3+2/3=1#