Re: Limit question

In article
<14456962.1196130891683.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
giuseppe <demarco.giu@xxxxxxxxx> wrote:

Anyone knows a method to solve this limit without
Hospital?

lim [ exp(x^2) - cos(x)]/(x^2)
x->0

Thanks

[exp(x^2) - cos(x)]/x^2 = [exp(x^2) - 1]/x^2 + [1 -
cos(x)]/(x^2). The
first term on the right -> 1 by definition of the
derivative. Multiply
the second term by (1+cos(x))/(1+cos(x)) to see its
limit is 1/2.

Yea but to simplify the second term you're assuming it is known lim[sin(x)/x] = 1. How do you prove that without L'Hospital (which effectively means without Taylor series, which effectively means without calculus)?
.