Re: Real Analysis Help ... Sequences

On Fri, 18 Apr 2008, Rob Johnson wrote:
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:

lim(n->oo) (sin n)/n = 0

because lim(n->oo) 1/n = 0 and |sin n| is bounded.

lim(n->oo) (sin 1/n) (1/n) = 1

because
lim(x->0) (sin x)/x = 1.

That is calculus.

That is circular, at least if you are thinking of using L'Hospital.
Taking the derivative of sin(x) requires evaluation of the limit of
sin(x)/x:

The reason it's calculus is that limit first occures calculating
d(sin x)/dx = cos x

The proof that lim_{x->0} sin(x)/x = 1 is geometric.

Though it's mostly a geometric proof, it isn't part of geometry.

sin(x) <= x <= tan(x)

Divide by sin(x) and take reciprocals to get

sin(x)
cos(x) <= ------ <= 1
x

Ok, how about an analytic proof? ;-)
Does that require defining sin and cos as solutions to
(D^2 + 1)y = 0 ?


Rob Johnson <rob@xxxxxxxxxxxxxx>
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