Q.Prove that the function $f(x) = x^n$ is continuous at $x = n$, where $n$ is a positive integer.

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

Step 1: Factor the difference of $n$th powers:

Step 2: Call the second factor $S(x) = x^{n-1} + x^{n-2}n + \cdots + n^{n-1}$. It is a sum of products of the continuous identity function $x$ with constants, hence $S(x)$ is continuous everywhere, so

Step 3: So $x^n = n^n + (x-n)S(x)$, and taking the limit as $x \to n$: …

Setting Up the Proof

To prove $f(x) = x^n$ is continuous at $x = n$ (for a positive integer $n$), we must show all three continuity conditions hold:

1. $f(n) = n^n$ is defined.
2. $\lim_{x \to n} f(x)$ exists.
3. $\lim_{x \to n} f(x) = f(n)$.

The first is immediate. The real work is showing $\lim_{x \to n} x^n = n^n$, and the cleanest in-syllabus way to do that is an algebraic factorisation, not an epsilon-delta chase.

Step-by-Step Reasoning

1. Factor $x^n - n^n$ using the standard difference-of-powers identity.

For any real $x$ and positive integer $n$:

This identity is exact and holds for every $x$, not just $x$ near $n$ — it comes from expanding the right-hand side and watching every middle term cancel in a telescoping pattern.

2. Name the second factor $S(x)$.

Let

Each term of $S(x)$ is a constant times a power of $x$, so $S(x)$ is built entirely from the continuous identity function $x$ using multiplication by constants and addition — by the algebra of continuous functions, $S(x)$ is continuous for every real $x$, in particular at $x = n$.

3. Evaluate $S(n)$.

Substituting $x = n$ into every term gives $n$ identical terms, each equal to $n^{n-1}$:

Since $S$ is continuous at $n$, $\lim_{x \to n} S(x) = S(n) = n^n$.

4. Rewrite $x^n$ using the factorisation and take the limit.

From Step 1, for every $x$:

Taking $x \to n$ on both sides, and using that a product of limits is the limit of the product (algebra of limits):

5. Conclude. …

Method: Proving $f(x)=x^n$ Is Continuous at $x=n$ Using Limit Laws

The standard Class 12 way to prove continuity of $f(x)=x^n$ at a point is an algebraic limit argument built on the factorisation of $x^n-n^n$ — not a formal epsilon-delta argument, which is outside the NCERT Class 12 syllabus for this chapter.

Steps

Step 1: State what must be shown

To prove continuity at $x=n$, it suffices to show

Step 2: Factor the difference $x^n-n^n$

Use the standard algebraic identity

valid for any positive integer $n$.

Step 3: Take the limit using the algebra-of-limits rules

Since each term $x^{n-1-k}n^{k}$ is itself a polynomial in $x$, the sum $x^{n-1}+x^{n-2}n+\cdots+n^{n-1}$ is continuous, so as $x\to n$ it tends to $n\cdot n^{n-1}=n^n$; and $(x-n)\to 0$. By the product rule for limits, …

Common Mistakes

Mistake 1: Trying to prove continuity using a formal epsilon-delta argument

Why it's wrong: NCERT Class 12 Continuity and Differentiability builds continuity proofs on the algebra of limits (limits of sums, products, and known limits of polynomials), not on a formal epsilon-delta construction — that is a university-level technique outside this syllabus and unnecessary complexity for what the algebraic factorisation already proves cleanly. Correct approach: factor $x^n-n^n=(x-n)(x^{n-1}+x^{n-2}n+\cdots+n^{n-1})$ and take the limit using the standard limit laws for sums and products.

Mistake 2: Misremembering the factorisation of $x^n-n^n$ …