How do you find the limit of #(sin2alpha)/sinalpha# as #alpha->0#?

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Answer 1 · 2016-10-14T02:36:36

The limit is #2#.

We can start by simplifying this expression using the double angle formula #sin2alpha = 2sinalphacosalpha#.

#=lim_(alpha ->0) (2sinalphacosalpha)/sinalpha#

#=lim_(alpha ->0) 2cosalpha#

You can substitute directly.

#=2cos(0)#

#=2(1)#

#=2#

Hopefully this helps!