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# Question:Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} \dfrac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

Last updated date: 02nd Aug 2024
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Hint:

To determine the points of discontinuity, evaluate the limits of $f(x)$ as $x$ approaches the values where the definition of the function changes. In this case, you should check the limit as $x$ approaches $0$ from the left and from the right.

Step-by-step Solution:

First analyze the function f at each of the two points in its domain: $x = 0$ and $x \neq 0$.

Case 1: $x = 0$

At x=0, f(x)=0.

To check if f is continuous at x=0, we need to check if the following three conditions hold:

1. f(0) exists. Yes, f(0)=0 exists.

2. $lim_{x \to 0} f(x)$ exists.

Let's find the one-sided limits of f at x=0:

$lim_{x \to 0^{-}} f(x) = lim_{x \to 0^{-}} \dfrac{|x|}{x} = lim_{x \to 0^{-}} \dfrac{-x}{x} = -1$

$lim_{x \to 0^{+}} f(x) = lim_{x \to 0^{+}} \dfrac{|x|}{x} = lim_{x \to 0^{+}} \dfrac{x}{x} = 1$

Since the one-sided limits do not agree, $lim_{x \to 0} f(x)$ does not exist.

1. $lim_{x \to 0} f(x) = f(0)$. Since $lim_{x \to 0} f(x)$ does not exist, this condition is automatically not met.

Therefore, f is not continuous at x=0.

Case 2: $x \neq 0$

At any point $x \neq 0$, $f(x)= \dfrac{∣x∣}{x}$. To check if f is continuous at x, we need to check if the following three conditions hold:

1. f(x) exists. Yes, $f(x) = \dfrac{∣x∣}{x}$ exists for all $x \neq 0$.

2. $lim_{x \to c} f(x)$ exists. Let's find the one-sided limits of f at x=c:

$lim_{x \to c^{-}} f(x) = lim_{x \to c^{-}} \dfrac{∣x∣}{x} = \dfrac{-c}{c} = -1$

$lim_{x \to c^{+}} f(x) = lim_{x \to c^{+}} \dfrac{∣x∣}{x} = \dfrac{c}{c} = 1$

Since the one-sided limits agree, $lim$lim_{x \to c} f(x)$exists and is equal to$\dfrac{c}{c}$. 1.$lim_{x \to c} f(x) = f(c)$. Since$lim_{x \to c} f(x) = \dfrac{c}{c} = 1$and$f(c)=1$, this condition is met. Therefore, f is continuous at all points$x \neq 0$. Note: The function$f(x) = \dfrac{|x|}{x}$​ effectively determines the sign of$x$. When$x$is positive,$f(x)$is 1, when$x$is negative,$f(x)$is -1, and at$x = 0$, the function is explicitly defined to be 0. This type of function that gives the sign of a number is sometimes called the "signum" or "sign" function, although the usual definition of the signum function does not have a value at$x = 0$. In this specific case, the function is discontinuous at$x = 0\$ because the limits from the left and right don't match the value of the function at that point.