How do you find the limit of #sinx/(5x)# as #x->0#?

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Answer 1 · 2016-10-25T10:05:21

The limit is #1/5#

Straight forward #sinx/(5x)=0# when #x->0#

So we have to apply l'Hôpital's rule

The limit is #((sinx)')/((5x)')=cosx/5#

And limit as #x->0# of #cosx/5=cos0/5=1/5#

Answer 2 · 2016-10-25T14:26:01

Rewrite it so you can use the fundamental trigonometric limit #lim_(thetararr0)sintheta/theta = 1#

#sinx/(5x) = 1/5(sinx/x)#
So

#lim_(xrarr0)sinx/(5x) = lim_(xrarr0)1/5(sinx/x)#

# = 1/5 lim_(xrarr0)sinx/x#

# = 1/5(1) = 1/5#