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

2 Answers
Alexander
Jul 24, 2016

#lim_(x->0) sin(2x)/sin(3x) = 2/3#

Explanation:

This limit is indeterminate since direct substitution yields #0/0#, which means that we can apply L'Hospital's rule, which simply involves taking a derivative of the numerator and the denominator.

#lim_(x->0) sin(2x)/sin(3x) -> 0/0#, so applying L'Hospital's rule:

#lim_(x->0) (2cos(2x))/(3cos(3x)) = 2/3#

Graph of #sin(2x)/sin(3x)#:

graph{sin(2x)/sin(3x) [-3.015, 3.142, -0.607, 2.471]}

Eddie
Jul 24, 2016

2/3

Explanation:

alternative answer

#lim_(x->0) sin(2x)/sin(3x) #

#lim_(x->0) 2* sin(2x)/(2x) *1/3 (3x) / sin(3x) #

#lim_(x->0) 2* sin(2x)/(2x) * lim_(x->0) 1/3 (3x) / sin(3x) #

let #u = 2x, v = 3x#

#lim_(u->0) 2* sinu/(u) * lim_(v->0) 1/3 (v) / sin v #

#= 2 * 1/3 = 2/3#

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