Example: Find $\lim\limits_{ x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$
$ = \frac{\sqrt{x^2 +9} - 3}{x^2} \circ \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{x^2 + 9 - 9}{x^2 ( \sqrt{x^2 + 9} + 3)} = \frac{x^2}{x^2 ( \sqrt{x^2 + 9} + 3)} = \frac{1}{ \sqrt{x^2 + 9} + 3} $
Conjecture: $\lim\limits_{ x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2} = \frac{1}{6}$
Prove with Rules (you don’t have to do this every time, this is just to prove a point)
$\lim\limits_{ x \to 0} x = 0, \lim\limits_{ x \to 0} x^2 = 0, \lim\limits_{ x \to 0} x^2 + 9 = \lim\limits_{ x \to 0} x^2 + \lim\limits_{ x \to 0} 9 = 9,$
$\lim\limits_{ x \to 0} \sqrt{x^2 + 9} = \sqrt{\lim\limits_{ x \to 0} x^2 + 9} = \sqrt{9} = 3, $
$\lim\limits_{ x \to 0} (\sqrt{x^2 + 9} + 3) = \lim\limits_{ x \to 0} \sqrt{x^2 + 9} + \lim\limits_{x \to 0} 3 = 3 + 3 = 6$
$\lim\limits_{ x \to 0} \frac{1}{\sqrt{x^2 + 9} + 3} = \frac{1}{\lim\limits_{ x \to 0} (\sqrt{x^2 + 9} + 3)} = \frac{1}{6}$
there is a geometric proof for $ \lim\limits_{x \to 0} \frac{ \sin x}{x} $, but i have no idea how to display it in this format, so you’ll have to google it for now.