Q.Show that the function defined by $f(x) = \sin(x^2)$ is a continuous function.

Concept: Continuity of Compositions — the composition of two everywhere-continuous functions is continuous.

• $g(x)=x^2$ is a polynomial, so continuous on $\mathbb{R}$.
• $h(u)=\sin u$ is continuous on $\mathbb{R}$. …

Continuity is preserved under composition: if the inner function is continuous at a point and the outer function is continuous at the value the inner function takes there, the composite is continuous at that point too. This is a powerful shortcut — it avoids re-proving continuity from the definition for every new function built out of familiar pieces.

Step 1 — Identify the inner and outer functions.

Write $f(x) = h(g(x))$ with

Step 2 — Check continuity of the inner function.

$g(x)=x^2$ is a polynomial, and every polynomial is continuous on all of $\mathbb{R}$.

Step 3 — Check continuity of the outer function.

$h(u)=\sin u$ is continuous for every real $u$ (a standard result for the sine function).

Step 4 — Apply the composition rule.

Since $g$ is continuous at every $x=a\in\mathbb{R}$ and $h$ is continuous at every value $g(a)=a^2$ that $g$ can take, the composite $f=h\circ g$ is continuous at every $a\in\mathbb{R}$. …

Method: Proving Continuity of a Composite Function (e.g., $\sin(x^2)$) via the Composition Rule

This method applies whenever a function can be written as "one function fed into another" — an inner function followed by an outer function.

Steps

Step 1: Identify the inner function $g$ and the outer function $h$.

Write the given function as $f(x) = h(g(x))$. For example, with $f(x)=\sin(x^2)$: inner $g(x)=x^2$, outer $h(u)=\sin u$.

Step 2: Establish that the inner function $g$ is continuous (at the point, or on the whole domain in question).

Cite the appropriate known result — e.g. $g(x)=x^2$ is a polynomial, hence continuous everywhere.

Step 3: Establish that the outer function $h$ is continuous at the value $g(a)$ the inner function actually produces — not just "somewhere."

This is the step most often done carelessly: it is not enough that $h$ is continuous in general; it must be continuous specifically at the point $u=g(a)$ that $g$ maps $a$ to. Check that $g(a)$ actually lies in the domain where $h$'s continuity is known to hold. …

Common Mistakes

Mistake 1: Assuming a composite function must be discontinuous or badly behaved because it "oscillates fast" or "looks complicated."

Why it's wrong: how quickly a continuous function oscillates has nothing to do with whether it is continuous — continuity is a purely local, point-by-point property about limits matching function values, not about visual complexity. Correct approach: verify continuity purely via the composition rule, never from a visual impression of the graph.

Mistake 2: Checking that the outer function is continuous "in general" without confirming it at the specific value $g(a)$ produced by the inner function.

Why it's wrong: the composition theorem requires continuity of the outer function at the point $g(a)$, not merely somewhere in its domain — for functions like $\sqrt{\ }$ or $\log$, whose domains are restricted, this distinction is exactly where a proof can silently break down if $g(a)$ falls outside that domain. Correct approach: always explicitly state the value $g(a)$ and confirm the outer function is continuous there. …