Exercise 5.1 · Q4: $x^n$ is continuous at $x=n$

Basics you need

Solution

1. Read the question: one $n,$ used twice

   The letter $n$ is a fixed positive integer, and it plays two roles: it is the power in $x^n$ and the point we test. So the claim is a separate statement for each $n$⁠:

   • $n = 1$⁠: $f(x) = x$ is continuous at $x = 1$⁠;
   • $n = 2$⁠: $f(x) = x^2$ is continuous at $x = 2$⁠;
   • $n = 3$⁠: $f(x) = x^3$ is continuous at $x = 3$⁠; and so on.

   $x^n$ is defined for every real $x,$ so $x = n$ is in the domain, and the value there is

   $$f(n) = n^n.$$
   Mixing up the two roles is the main trap. Once $n$ is fixed, $n^n$ is just a number (for example $3^3 = 27$⁠), and the problem is an ordinary "continuity at a point" question.

2. Expand $f(n + h)$ and $f(n - h)$

   Let $h > 0.$ By the binomial theorem with $a = n$ and $b = h$⁠:

   $$\begin{aligned} (n + h)^n &= n^n + \sum_{k=1}^{n} \binom{n}{k} n^{n-k} h^k\\ &= n^n + n\cdot n^{n-1} h + \dots + h^n. \end{aligned}$$

   With $b = -h$ instead:

   $$(n - h)^n = n^n + \sum_{k=1}^{n} \binom{n}{k} n^{n-k} (-h)^k.$$
   Separating the $k = 0$ term from the rest is the key move. The first term $n^n$ does not involve $h$⁠; every remaining term has a factor $h^k$ with $k \ge 1,$ multiplied by the fixed number $\binom{n}{k} n^{n-k}.$
   Binomial theorem, Class XI. The sum from $k = 1$ to $n$ has exactly $n$ terms: a fixed, finite number.

3. Take the limits as $h \to 0$

   Each term of the form $\binom{n}{k} n^{n-k} (\pm h)^k$ with $k \ge 1$ is a constant times $(\pm h)^k,$ so it tends to 0. There are only $n$ such terms, so their sum tends to 0 (Basic 4). Therefore

   Left-hand limit.

   $$\lim_{x\to n^-} f(x) = \lim_{h\to0} (n - h)^n = n^n + 0 = n^n.$$

   Right-hand limit.

   $$\lim_{x\to n^+} f(x) = \lim_{h\to0} (n + h)^n = n^n + 0 = n^n.$$
   "Finite" matters: $n$ is fixed before $h$ moves, so the number of terms does not grow as $h$ shrinks, and the limit of a sum of $n$ terms is the sum of their limits.
   Basics: left-hand and right-hand limits

4. Compare the three numbers

   $$\lim_{x\to n^-} f(x) = n^n,\quad \lim_{x\to n^+} f(x) = n^n,\quad f(n) = n^n.$$

   All three agree, so $f(x) = x^n$ is continuous at $x = n$, for every positive integer $n.$

   This is Definition 1 at the point $c = n.$ Nothing in Steps 2 and 3 used that the point equals the exponent: replacing the point $n$ by any real $c$ gives $(c \pm h)^n \to c^n.$ So $x^n$ is in fact continuous at every real number.
   Lesson: continuity at a point

5. See it for $n = 2$ and $n = 3$⁠; the shortcut

   For $n = 2$ (testing $x^2$ at 2): $(2 \pm h)^2 = 4 \pm 4h + h^2 \to 4 = f(2).$

   For $n = 3$ (testing $x^3$ at 3):

   $$(3 \pm h)^3 = 27 \pm 27h + 9h^2 \pm h^3 \to 27.$$

   The shortcut: $x^n$ is a polynomial, and Example 14 proves every polynomial is continuous at every real number. Equivalently, $x^n = x \cdot x \cdots x$ ($n$ factors), the identity function $x$ is continuous (Example 6), and a product of continuous functions is continuous (Theorem 1 (3)), applied $n - 1$ times.

   The concrete cases show the general expansion at work: the middle coefficients $\binom{3}{1}3^2 = 27$ and $\binom{3}{2}3 = 9$ are fixed numbers multiplying powers of $h.$ The shortcut is a valid proof too; the first-principles route above shows why it is true.
   Theorem 1 (3), Section 5.2.1: $f \cdot g$ is continuous at $c$ if $f$ and $g$ are. Lesson: algebra of continuous functions

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