Dear Uncle Colin,

Why is the limit of $\frac{\sin(2x)}{x}$ and $\frac{\sin(x)}{x}$ different, as $x\to 0$? Changing the argument of $\sin$ only changes the period of the function, so how come the value changes as well?

- Limits Hitting Origin: Pain In The Arsinh, LOL.

Hi, LHOPITAL, and thanks for your message!

Indeed, as $x$ approaches 0, $\frac{\sin(x)}{x}$ approaches 1 and $\frac{\sin(2x)}{x}$ approaches 2. There are a couple of ways I might approach this: one slightly mechanical, and one a bit deeper.

### The mechanical way

$\sin(2x)$ is the same as $2\sin(x)\cos(x)$, so $\frac{\sin(2x)}{x} = \frac{2\sin(x)\cos(x)}{x}$. As $x \to 0$, $2\cos(x)$ approaches 2, so the value of the whole thing is close to $\frac{2 \sin(x)}{x}$, which approaches double the other limit.

### The deeper way

It’s not quite true that changing the argument just changes the period of the function: it changes its gradient as well. The gradient of $\sin(2x)$ is double that of $\sin(x)$. Since we’re dividing by $x$, it’s the ratio of gradients that’s important – and because $\sin(2x)$ is steeper, it gives a higher value in the limit.

Hope that helps!

- Uncle Colin