Exercise 5.1 · Q23: $\frac{\sin x}{x}$ for $x\lt0$⁠, $x+1$ for $x\ge0$⁠: any breaks?

Basics you need

Solution

1. The domain, split into three parts

   For $x\lt0$ the rule $\frac{\sin x}{x}$ is defined because $x\ne0$⁠. For $x\ge0$ the rule $x+1$ is defined. So $f$ is defined at every real number. We examine three kinds of point: $c\lt0$⁠, $c\gt0$⁠, and the junction $c=0$⁠.

   "Find all points of discontinuity" asks about every point of the domain, here all of $\R$⁠. The interiors of the two pieces are handled by known theorems; the junction needs a direct check.
   Splitting the real line at the junction, as in Examples 10 and 13. Basics: piecewise functions

2. Every $c\lt0$ is a point of continuity

   Fix $c\lt0$⁠. Every $x$ in the open interval $(-\infty,0)$⁠, which contains $c$⁠, has $f(x)=\frac{\sin x}{x}$⁠. On that interval $\sin x$ and $x$ are continuous and the denominator is never $0$⁠, so by Theorem 1 (4)

   $$\lim_{x\to c}f(x)=\lim_{x\to c}\frac{\sin x}{x}=\frac{\sin c}{c}=f(c).$$
   The limit at $c$ only uses $x$ close to $c$⁠, and those $x$ are all negative, so only the first rule is involved. The condition $c\ne0$ of Theorem 1 (4) holds because $c\lt0$⁠.
   Theorem 1 (4); sine continuous (Example 17); identity continuous (Example 6). Lesson: algebra of continuous functions

3. Every $c\gt0$ is a point of continuity

   Fix $c\gt0$⁠. On the open interval $(0,\infty)$ around $c$⁠, $f(x)=x+1$⁠, a polynomial. So

   $$\lim_{x\to c}f(x)=c+1=f(c).$$
   Polynomials are continuous everywhere (Example 14), and near $c\gt0$ the function coincides with one.

4. At the junction: the value $f(0)$

   $0$ satisfies "$x\ge0$", so the second rule applies:

   $$f(0)=0+1=1.$$
   The expression $\frac{\sin 0}{0}$ is meaningless, but it never arises: the definition of $f$ sends $0$ to the rule $x+1$⁠.

5. At the junction: the left-hand limit

   For $x\lt0$⁠, $f(x)=\frac{\sin x}{x}$⁠, so

   $$\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{\sin x}{x}.$$

   Write $x=-t$ with $t\gt0$⁠, $t\to0$⁠. Since $\sin(-t)=-\sin t$⁠,

   $$\frac{\sin x}{x}=\frac{-\sin t}{-t}=\frac{\sin t}{t}\ \longrightarrow\ 1.$$

   So $\lim_{x\to0^-}f(x)=1$⁠.

   The standard limit (Basic 3) holds as $x\to0$ from either side; the substitution shows explicitly that the left side gives the same values as the right side, because $\frac{\sin x}{x}$ is an even function. A table makes it concrete: at $x=-0.1,\,-0.01,\,-0.001$ the values are $0.9983342$⁠, $0.9999833$⁠, $0.9999998$⁠.
   $\lim_{t\to0}\frac{\sin t}{t}=1$⁠, Class XI, Chapter 12. Basics: what a limit is

6. At the junction: the right-hand limit

   For $x\gt0$⁠, $f(x)=x+1$⁠. With $x=h$⁠, $h\gt0$⁠, $h\to0$⁠:

   $$\lim_{x\to0^+}f(x)=\lim_{h\to0}\,(h+1)=1.$$

7. Compare and conclude

   From Steps 4 to 6:

   $$\begin{aligned} \lim_{x\to0^-}f(x)&=1,\\ f(0)&=1,\\ \lim_{x\to0^+}f(x)&=1. \end{aligned}$$

   All three are equal, so $f$ is continuous at $0$⁠. Together with Steps 2 and 3, $f$ is continuous at every real number, and $f$ has no point of discontinuity.

   Every point of the domain falls into exactly one of the three cases, and in each case Definition 1 is satisfied. So the set of points of discontinuity is empty.
   Continuity at a junction. Lesson: continuity at a point

Every fact used, and where it comes from