What is the Limit h->0 if

Sin2h/sinh ???

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- Oct 2nd 2006, 06:12 AMLuckyjoshuaLimits and sin
What is the Limit h->0 if

Sin2h/sinh ??? - Oct 2nd 2006, 06:21 AMCaptainBlack
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 PMThePerfectHacker