How do you evaluate the limit #sin(3t)/(5t)# as t approaches #0#?

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Answer 1 · 2016-08-06T13:14:12

#3/5#

we'll shoe horn this into well known limit #lim_{z to 0} (sin z)/z = 1#

so

#lim_{t to 0} (sin (3t))/(5t)#

#=lim_{t to 0} 3/5 (sin (3t))/(3t)#

let #z = 3t#

#=lim_{z to 0} 3/5 (sin (z))/(z)#

#= 3/5lim_{z to 0} (sin (z))/(z) = 3/5#