Find the limit of (sin x)/x as x--> 0
For simplicity
sinx can be expanded as
$\displaystyle x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}........$
So the Limit now becomes
$\displaystyle
Lim_{x->0} \frac{x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}........}{x}
$
$\displaystyle
= Lim_{x->0} {1 -\frac{x^3}{x*3!} + \frac{x^5}{x*5!} - \frac{x^7}{x*7!}........}
$
$\displaystyle
= 1
$
Its better if you follow Chris
Or you can use the definition of the derivative :
$\displaystyle \lim_{x \to 0} \frac{f(a+x)-f(a)}{x}=f'(a)$
let f be the sine function, and a=0, remember that the derivative of the sine function is cosine, and observe
I guess this method suits the most what you're currently doing...