limit

**ah-bee** · May 8th 2008, 05:58 AM

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.

**wingless** · May 8th 2008, 06:08 AM

$\lim_{x\to 0}\frac{\sin kx}{x}$

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

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

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

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

$1\cdot k = k$

**Isomorphism** · May 8th 2008, 06:12 AM

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: $\lim_{x \to 0} \frac{\sin(kx)}{x} = k\lim_{x \to 0} \frac{\sin(kx)}{kx}$

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