# Limits and sin

• Oct 2nd 2006, 06:12 AM
Luckyjoshua
Limits and sin
What is the Limit h->0 if

Sin2h/sinh ???
• Oct 2nd 2006, 06:21 AM
CaptainBlack
Quote:

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
• Oct 2nd 2006, 04:11 PM
ThePerfectHacker
Quote:

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