Thread: Limits and sin

What is the Limit h->0 if

Sin2h/sinh ???

Originally Posted by Luckyjoshua

What is the Limit h->0 if

Sin2h/sinh ???

for small x sin(x)~x, so

sin(2h)/sin(h)~2h/h=2, so lim(sin(2h)/sin(h), h ->0)=2

This is an informal argument (but is the right answer) if you need this
done formally one good route is L'Hopitals rule. Using this we have:

lim(sin(2h)/sin(h), h ->0)=lim([d(sin(2h)/dh]/[d(sin(h)]dh, h ->0)

.................................=lim([2 cos(2h)]/[cos(h)], h ->0)

.................................=lim([2 cos(2h)], h ->0)/lim([cos(h)], h ->0)=2

RonL

Originally Posted by CaptainBlack

This is an informal argument (but is the right answer) if you need this
done formally one good route is L'Hopitals rule.

Watch und learn, furm die Meister:

Express as,
sin 2x/sin x=[(sin 2x)/x]/[sin x/x]=2[sin 2x/(2x)]/[sin x/x]
Now,
lim sin2x/2x=lim sin x/x =1 (limit composition rule)
lim sin x/x=1
Thus,
2(1)/(1)=2