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