Limit question..

[zuuberbat]

Don't really understand how to do these..

1) Use the fact that lim h->0 ((e^h - 1) / h) = 1 to show that lim h->0 ((e^h cosh - 1) / h) = 1

2) Use the fact that lim h->0 ((e^h cosh - 1) / h) = 1 to determine the derivative of f(x) = e^x sinx

F exam coming up soon; if you can please provide explanations/steps, thanks a bunch.

 

[arthur ohlsten]

h-->0 [e^h cosh -1]/ h = ?
h-->0 [[1+h + h^2/2 +....]cos h -1] /h
but cosh h-->0=1
h-->0 [1+h+h^2/2 +...-1]/h
h-->0 [h+h^2/2 + ....]/h
h-->0=1+h/2+h^2/3+...
h-->0=1

====================================================================
f[x]= e^x sin x
f[x+h]= e^(x+h) sin [x+h]
f[x+h= e^x e^h [sinxcosh +cosx sin h]
f[x+h]-f[x]= e^x e^h [sinxcosh+cosxsinh]- e^x sin x
f[x+h]-f[x]= e^xsinx [e^hcosh -1] +e^x e^h cosx sin h
[f[x+h]-f[x]]/h =e^x sin x [e^h cos h -1] /h + e^x e^h cosx sinh/h

but h-->0
e^h cos h -1 =0
sin h=h
e^h=1
[ f[x+h]-f[x]]/h = f ' [x]
f ' [x] = e^x cos x answer

Arthur