# identifying discontinuity

In the following function, how do you qualitatively show that the function is discontinuous?

$$f(x) = \frac{\sqrt{1 - \sqrt{\sin(2x)}}}{\pi - 4x}$$

I started off by analyzing the numerator and denominator separately. The maximum value of $\sin$ is one. The factor $2$ in the sin expression makes the $\sin$ graph look thinner (the period decreases by a factor of $2$). The maximum value of $\sin$ is attained at $\pi\over 2$. At $\pi \over 2$, the numerator goes to zero. In the vicinity of $\pi \over 2$, the $\sin$ function does not spit out negative values. Whether you approach from the left side or from the right side, the numerator as a whole approaches zero.

The denominator is positive as you approach from the left-hand side and negative as you approach from the right-hand side. Therefore, there is a sign flip.

To decide if the given function is continuous at $\pi \over 4$, I need to know if the limit on both the sides agree upon one value (the change of sign is irrelevant if they hit zero).

What do I do now?

Original question

The actual question was to calculate the following limit:

$$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{1 - \sqrt{\sin(2x)}}}{\pi - 4x}$$

The answer was "does not exist"; left-hand limit is not equal to the right-hand limit.

Graph of the function:

## 1 Answer

Hint:

You can write:

$$\sqrt{1-\sqrt{\sin(2x)}}=\frac{\sqrt{1-\sin(2x)}}{\sqrt{1+\sqrt{\sin(2x)}}}=\frac{|\cos(2x)|}{\sqrt{1+\sqrt{\sin(2x)}}\sqrt{1+\sin(2x)}}$$