Q.Prove that the identity function on real numbers given by $f(x) = x$ is continuous at every real number.

Concept: Continuity At A Point — $f$ is continuous at $x=a$ if $\displaystyle\lim_{x\to a}f(x)=f(a)$.

For $f(x)=x$: $f(a)=a$, and by the direct limit law, $\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}x=a$. …

To prove continuity at a point $x=a$, the NCERT test is to check three conditions: $f(a)$ is defined, $\displaystyle\lim_{x\to a}f(x)$ exists, and this limit equals $f(a)$. For the identity function this is immediate because the function's rule is the input itself.

Step 1 — Take an arbitrary real number $a$.

Since $a$ was not chosen to be special, showing continuity at this generic $a$ proves continuity at every real number.

Step 2 — Confirm $f(a)$ is defined.

$f(a) = a$, a real number, for every $a \in \mathbb{R}$ — the identity function has domain all of $\mathbb{R}$.

Step 3 — Evaluate the limit.

By the standard limit law for the identity function, $\displaystyle\lim_{x\to a} f(x) = \lim_{x\to a} x = a$. The limit as $x$ approaches $a$ of the function "$x$" is simply $a$ — there is no algebraic manipulation needed, since $f$ does not distort its input at all.

Step 4 — Compare the limit with $f(a)$.

All three continuity conditions hold.

Step 5 — Conclude for every point. …

Method: Proving Continuity Using the ε–δ Definition

This method is the rigorous, formal way to prove continuity at a point (or at every point) when the proof is expected to be backed by the precise limit definition rather than just quoted from limit laws.

Steps

Step 1: State the ε–δ definition to be satisfied

This must be shown to hold for the chosen point $a$ (or for an arbitrary $a$, to prove continuity everywhere).

Step 2: Express $|f(x) - f(a)|$ in terms of $|x-a|$

Algebraically simplify the output-distance expression so that it is bounded above by some expression involving $|x - a|$. For simple functions this simplification is often exact and immediate; for more complex ones it may require an inequality bound.

Step 3: Choose $\delta$ in terms of $\varepsilon$ …

Common Mistakes

Mistake 1: Treating the geometric "unbroken graph" argument as a substitute for the formal proof

Why it's wrong: saying "the graph of $y=x$ is a straight line, so obviously it's continuous" is intuitively true but is not a proof by the $\varepsilon$–$\delta$ definition, which is specifically what this type of question demands. Correct approach: always set up the formal statement — for every $\varepsilon>0$, exhibit a valid $\delta>0$ — even when the geometric picture already seems obvious.

Mistake 2: Choosing $\delta$ before simplifying $|f(x)-f(a)|$

Why it's wrong: the correct choice of $\delta$ must come from the simplified expression for $|f(x)-f(a)|$ in terms of $|x-a|$; guessing a fixed $\delta$ first (e.g. $\delta=1$) usually fails to guarantee the required bound for every $\varepsilon$. Correct approach: simplify $|f(x)-f(a)|$ first, see exactly how it relates to $|x-a|$, and only then choose $\delta$ (here $\delta=\varepsilon$ works because the two distances are literally equal). …