How do you evaluate the limit #sin(ax)/sin(bx)# as x approaches #0#?

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Eddie Share
Sep 2, 2016

Answer:

# = a/b#

Explanation:

#lim_(x to 0) sin(ax)/sin(bx)# is in #0/0# indeterminate form so we can use l'Hopital's rule

#= lim_(x to 0) (a cos (ax))/(b cos(bx))#

we can lift out the contant term and note that the limit of the quotient is the quotient of the limits where the limits are known

#=a/b (lim_(x to 0) cos (ax))/(lim_(x to 0) cos(bx))#

#=a/b lim_(x to 0) (1)/(1) = a/b#

Another approach is to head for known limit #lim_(z to 0) (sin z)/z = 1# and to proceed as follows

#lim_(x to 0) sin(ax)/sin(bx)#

#=lim_(x to 0) sin(ax)/ x x / (sin(bx))#

#=lim_(x to 0) (sin(ax)/ x)/( (sin(bx)) /x )#

#=a/b lim_(x to 0) (sin(ax)/ (ax))/( (sin(bx)) /(bx) )#

#=a/b (lim_(x to 0) sin(ax)/ (ax))/(lim_(x to 0) (sin(bx)) /(bx) )#

with subs #u = ax, v = wz#

#=a/b (lim_(u to 0) sin(u)/ (u))/(lim_(v to 0) (sin(v)) /(v) )#

#=a/b *1/1#

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