(without using l'hopital's rule) how to do you prove that (sinx)/x=1?

For the second one, we have

tan(3x)/x  = (sin3x)/x * 1/cos3x = (3sin3x/3x) * 1/(cos 3x) 

Let 3x = Θ

And  using the fact that sinΘ/Θ = 1, and cosΘ = 1   as Θ→ 0..... we have

(3sinΘ/Θ) * 1/cosΘ = (3)(sinΘ/Θ) * 1/cosΘ =

[3(1)] / [1/1]  = 3/1  = 3  

(without using l'hopital's rule) how to do you prove that:

lim x->0 (sinx)/x=1 ? 

$$\small{\text{ $ \boxed{ \lim\limits_{x\to0} \left( \dfrac{ \sin{(x)} } { x } \right) \qquad \sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\dotsb } % boxed $ }}$\\\\\\$ \small{\text{ $ \lim\limits_{x\to0}\left( \dfrac{ \sin{(x)} } { x } \right) = \lim\limits_{x\to0}\left( \dfrac{ \dfrac{x}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\mp\dotsb } { x } \right) = \lim\limits_{x\to0}\left( \dfrac{ x \left( \dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right) } { x } \right) $ }}$\\\\\\$ \small{\text{ $ = \lim\limits_{x\to0}\left( \dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)= \dfrac{1}{1!} = \textcolor[rgb]{1,0,0}{1} $ }}$$

(without using l'hopital's rule) how to do you prove that:

lim x->0 (tan3x)/x =3 ? 

$$\small{\text{ $ \boxed{ \lim\limits_{x\to0} \left( \dfrac{ \tan{(3x)} } { x } \right) \qquad \tan{ (3x) } &= (3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb } % boxed $ }}$\\\\\\$ \small{\text{ $ \lim\limits_{x\to0}\left( \dfrac{ \tan{(3x)} } { x } \right) = \lim\limits_{x\to0}\left( \dfrac{ (3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb } { x } \right) = \lim\limits_{x\to0}\left( \dfrac{ x \left( 3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right) } { x } \right) $ }}$\\\\\\$ \small{\text{ $ = \lim\limits_{x\to0}\left( 3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)= \textcolor[rgb]{1,0,0}{3} $ }}$$