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

1 Answer
Jul 1, 2016

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

Explanation:

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