At x = 0, the function $${\text{f}}\left( {\text{x}} \right) = \left| {\sin \frac{{2\pi {\text{x}}}}{{\text{L}}}} \right|$$ , (-$$\infty $$ < x < $$\infty $$, L > 0) is A. continuous and differentiable B. Not continuous and not differentiable C. Not continuous but differentiable D. Continuous but not differentiable

The correct answer is: D. Continuous but not differentiable

The function $f(x) = | \sin \frac{2\pi x}{L} | $ is continuous at $x=0$ because the two-sided limit $\lim_{x\to 0} f(x)$ exists and is equal to $f(0)$. However, the function is not differentiable at $x=0$ because the derivative $f'(x)$ does not exist at $x=0$.

To see why the derivative does not exist at $x=0$, we can consider the following graph of the function:

[asy]
unitsize(1 cm);

draw((0,-1.2)–(0,1.2));
draw((-2,0)–(2,0));

real g(real x) {
return abs(sin(2pix/1));
}

draw(graph(g,-2,2),red);

label(“$x$”, (2,0), E);
label(“$y$”, (0,1.2), N);
[/asy]

As you can see, the graph of the function has a sharp point at $x=0$. This means that the function is not differentiable at $x=0$.

In general, a function is differentiable at a point $x=a$ if and only if the two-sided limit $\lim_{x\to a} \frac{f(x) – f(a)}{x-a}$ exists and is equal to $f'(a)$. In the case of the function $f(x) = | \sin \frac{2\pi x}{L} | $, the two-sided limit $\lim_{x\to 0} \frac{f(x) – f(0)}{x-0}$ does not exist, so the function is not differentiable at $x=0$.