How do you find the limit of #(Sin5x) / 5# as x approaches 0?

2 Answers
Konstantinos Michailidis
May 18, 2016

We know that #lim_(x->0) sinx/x=1# hence

#lim_(x->0) x*sin(5x)/(5x)=lim_(x->0)x*(lim_(x->0) sin(5x)/(5x))= 0*1=0#

mason m
May 18, 2016

#0#

Explanation:

We can plug in #0# straightaway, since it doesn't cause any domain errors:

#lim_(xrarr0)sin(5x)/5=sin(5*0)/5=sin0/5=0#

Examining the graph of #sin(5x)/5# reveals the point #(0,0)# is on the function:

graph{sin(5x)/5 [-1.2, 1.2, -0.599, 0.601]}

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