Q.Examine that $\sin |x|$ is a continuous function.

The key idea is that $\sin |x|$ is a composition of the continuous absolute value function $|x|$ and the continuous sine function $\sin t$, and the composition of continuous functions is continuous.

Step 1: The absolute value function $f(x) = |x|$ is continuous for all real $x$ (it is a piecewise linear function with no jumps).

Step 2: The sine function $g(t) = \sin t$ is continuous for all real $t$ (its graph is a smooth wave). …

The key idea here is continuity of compositions. If you have two functions $f$ and $g$, and $g$ is continuous at a point $c$, and $f$ is continuous at $g(c)$, then the composite function $f(g(x))$ is continuous at $c$. This is a theorem that holds for all real numbers.

Now, $\sin |x|$ is exactly that: take $g(x) = |x|$ (the inside function) and $f(u) = \sin u$ (the outside function). So $\sin |x| = f(g(x))$.

Let's check each piece:

1. The absolute value function $g(x) = |x|$ is continuous everywhere on $\mathbb{R}$. You know this from its graph — it's a V-shape with no jumps or breaks. At $x=0$, the left and right limits both equal $0$, and $|0| = 0$, so it's continuous there too.

2. The sine function $f(u) = \sin u$ is continuous for all real $u$. This is a standard result from trigonometry — sine is smooth and periodic, with no discontinuities anywhere.

3. Composition rule: Since $g$ is continuous at every $x$, and $f$ is continuous at every $u = g(x)$, the composition $f(g(x)) = \sin |x|$ is continuous at every $x \in \mathbb{R}$. …

Method: Composition Proofs with an Absolute Value as the Inner Function

Here the roles are reversed from a case like $|\cos x|$: the absolute value is the inner function, and the corner it creates sits inside a smooth outer function.

Steps

Step 1: Identify inner and outer functions

Write the function as $h(g(x))$ with $g(x) = |x|$ (the inner function) and $h$ the outer standard function applied to it.

Step 2: Confirm the inner function $|x|$ is continuous for every real $x$, including at its corner point $x=0$

Check the left-hand limit, right-hand limit, and the value at $0$ all agree (all equal $0$) — the corner affects differentiability there, never continuity.

Step 3: Confirm the outer function is continuous on the entire range $|x|$ can produce …

Common Mistakes

Mistake 1: Assuming the corner in $|x|$ at $x=0$ makes $\sin|x|$ discontinuous there

Why it's wrong: $|x|$ is continuous at $x=0$ (the left-hand limit, right-hand limit, and value all equal $0$) even though it is not differentiable there — continuity and differentiability are separate properties, and only the former is being asked about. Correct approach: check the three continuity conditions directly at $x=0$ (or trust the composition theorem, since $|x|$ is continuous everywhere) rather than reasoning from the sharp corner.

Mistake 2: Skipping the domain check on the outer function …