Exercise 5.1 · Q24: Is $x^2\sin\frac1x$ (with $f(0)=0$⁠) continuous?

Basics you need

Solution

1. The domain

   For $x\ne0$⁠, $\frac1x$ is defined, so $x^2\sin\frac1x$ is defined; and $f(0)=0$ is given. So $f$ is defined on all of $\R$⁠. We check continuity at every $c\ne0$⁠, then at $0$⁠.

2. Every $c\ne0$ is a point of continuity

   Fix $c\ne0$⁠. Build $x^2\sin\frac1x$ step by step:

   Inner function $h(x)=\frac1x$⁠: a rational function with non-zero denominator at $c$⁠, so continuous at $c$ (Example 16, or Example 9).

   Outer function $g(u)=\sin u$⁠: continuous at every real $u$⁠, in particular at $u=\frac1c$ (Example 17). By Theorem 2, $\sin\frac1x$ is continuous at $c$⁠.

   Product: $x^2$ is a polynomial, continuous at $c$ (Example 14). By Theorem 1 (3), $x^2\cdot\sin\frac1x$ is continuous at $c$⁠.

   Finally, every $x$ with $|x-c|\lt|c|$ has the same sign as $c$⁠, so $x\ne0$ and $f(x)=x^2\sin\frac1x$ on that open interval around $c$⁠. Hence

   $$\lim_{x\to c}f(x)=c^2\sin\frac1c=f(c).$$
   The special value at $0$ does not interfere: the interval $|x-c|\lt|c|$ stays on one side of $0$⁠, so near $c$ the function is given by the single formula. The chain "inner, outer, then product" mirrors exactly how the formula is built.
   Theorem 2 (composites) and Theorem 1 (3) (products). Lesson: algebra of continuous functions

3. The value at $0$

   $$f(0)=0\quad\text{(given)}.$$
   The formula $x^2\sin\frac1x$ cannot be used at $0$ ($\frac10$ is undefined), which is why the question supplies the value separately.

4. The key bound

   For every $x\ne0$⁠, $\left|\sin\frac1x\right|\le1$ (Basic 3). Multiplying by $x^2\ge0$⁠:

   $$|f(x)|=x^2\left|\sin\tfrac1x\right|\le x^2,$$

   that is,

   $$-x^2\le f(x)\le x^2\qquad(x\ne0).$$
   We cannot control $\sin\frac1x$ itself near $0$⁠, but we can control its size: it is never bigger than $1$⁠. Multiplying an inequality by a non-negative number keeps its direction, so the bound passes to the product.
   $|ab|=|a|\,|b|$ and $|x^2|=x^2$⁠.

5. Both one-sided limits at $0$

   Right-hand limit. For $x\gt0$⁠, $-x^2\le f(x)\le x^2$⁠, and $\lim_{x\to0^+}(-x^2)=0$ $=\lim_{x\to0^+}x^2$⁠. By the sandwich theorem,

   $$\lim_{x\to0^+}f(x)=0.$$

   Left-hand limit. For $x\lt0$ the same bound holds (it was proved for every $x\ne0$⁠), and $\pm x^2\to0$ as $x\to0^-$⁠. So

   $$\lim_{x\to0^-}f(x)=0.$$
   The function is trapped between two walls that pinch together at the origin, from both sides. Whatever the wiggles do, they cannot escape the shrinking gap between $-x^2$ and $x^2$⁠.
   Sandwich theorem, Class XI, Chapter 12. Basics: what a limit is

6. Compare

   $$\begin{aligned} \lim_{x\to0^-}f(x)&=0,\\ f(0)&=0,\\ \lim_{x\to0^+}f(x)&=0. \end{aligned}$$

   All three agree, so $f$ is continuous at $0$⁠. With Step 2, $f$ is continuous at every real number: yes, $f$ is a continuous function.

   Continuity at a point. Lesson: continuity at a point

7. Why the factor $x^2$ is essential

   Without it, $\sin\frac1x$ has no limit at $0$⁠. Take $x_k=\frac{1}{\frac{\pi}{2}+2k\pi}$ and $y_k=\frac{1}{\frac{3\pi}{2}+2k\pi}$⁠, $k=1,2,3,\ldots$ Both sequences tend to $0$ from the right, yet

   $$\begin{aligned} \sin\frac{1}{x_k}&=\sin\left(\tfrac{\pi}{2}+2k\pi\right)=1,\\ \sin\frac{1}{y_k}&=\sin\left(\tfrac{3\pi}{2}+2k\pi\right)=-1. \end{aligned}$$

   Arbitrarily close to $0$ the values $1$ and $-1$ both occur, so they cannot settle on one number. With the factor $x^2$⁠, those values become $x_k^2$ and $-y_k^2$⁠, both tending to $0$⁠.

   This explains the question's interest: a wildly oscillating factor can still give a continuous function, provided it is multiplied by something that tends to $0$ and the oscillation stays bounded.

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