1. ## limit

I have always taken for granted sin, cos, and tan... but now i want to try to understand something about the unit circle which i have neglected through high school... lol. Well, I have a statement that says sin x < x < tan x. Can someone please bring me through why this is the case geometrically. Also, i am able to prove lim(x->0) sin(x)/(x) = 1 from this but just wondering what the proof for lim(x->0) sin(kx)/(x) = k looks like.

2. $\displaystyle \lim_{x\to 0}\frac{\sin kx}{x}$

$\displaystyle \lim_{x\to 0}\frac{\sin kx}{x}\cdot \frac{k}{k}$

$\displaystyle \lim_{x\to 0}\frac{\sin kx}{kx}\cdot k$

When x approaches zero, kx will approach zero too. So we can write it as,

$\displaystyle \lim_{kx\to 0}\underbrace{\frac{\sin kx}{kx}}_{\text{This makes 1}}\cdot k$

$\displaystyle 1\cdot k = k$

3. Originally Posted by ah-bee
I have always taken for granted sin, cos, and tan... but now i want to try to understand something about the unit circle which i have neglected through high school... lol. Well, I have a statement that says sin x < x < tan x. Can someone please bring me through why this is the case geometrically.
Try this PlanetMath: comparison of $\sin þeta$ and $þeta$ near $þeta = 0$

Originally Posted by ah-bee
Also, i am able to prove lim(x->0) sin(x)/(x) = 1 from this but just wondering what the proof for lim(x->0) sin(kx)/(x) = k looks like.
Hint: $\displaystyle \lim_{x \to 0} \frac{\sin(kx)}{x} = k\lim_{x \to 0} \frac{\sin(kx)}{kx}$