about Hyperbolic Trigonometry !

答非所問!!!

luie3e has misinterpret that sinh(x) = sin(hx)
and the questioner even didn't recognize it.

sinh(x) = [e^x - e^(-x)]/2 as the other answerers suggested.

But the questioner choose the ONLY wrong one.

lim (sinh x) = lim {1/2*[e^x - e^(-x)]}
= 1/2 * lim[e^x - e^(-x)]
As x tends to 0, lim e^x = 1, lim [e^(-x)] = 1
So lim (sinh x) = 1/2 * (1 - 1) = 0

lim x as x-> 0 = 0

ONLY BOTH lim (sinh x) AND lim x HAVE BEEN SHOWN TO BE ZERO, CAN L'HOSPITAL'S RULE BE APPLIED TO FIND lim (sinh x / x) as x -> 0.

So as x -> 0,
lim (sinh x / x) = lim [(d(sinh x)/dx) / (dx/dx)]
= lim cosh x
= 1/2 lim [e^x + e^(-x)]
= 1/2 * (1 + 1)
= 1.

Obvious, this is NOT because of lim (sin x / x) or lim (sinc x) = 1 as x tends to zero.

2006-12-13 22:09:15 補充：
First, there is a grammatical mistake in the last sentence. It should be "OBVIOUSLY, this is NOT because of...".

2006-12-13 22:10:26 補充：
Second, I would like to point out that before using L-Hospital's rule, we must show that both denominator and nominator tend to all zero or all infinity before it can be applied. Otherwise, it is NOT true.

2006-12-13 22:51:08 補充：
如果呢條係考試題目, 欠缺證明 lim sinh x = 0 同埋 lim x = 0而直接用L'Hospital's rule係會被扣分的.因為要必須符合這些先決條件, L'Hospital's rule才是成立. 用方程式必須要瞭解應用的條件和局限, 才能恰當地使用.

sinh(x)=(e^x-e^(-x))/2
lim(x->0)[sinh x/ x]
=lim(x->0)[(e^x-e^(-x))/2x]
=lim(x->0)[(e^x+e^(-x))/2]
=(e^0+e^(0))/2
=(1+1)/2
=1