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!

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