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# Question:Find all points of discontinuity of $f$, where $f$ is defined by:$f(x) = \begin{cases} \dfrac{x}{|x|} & \text{if } x < 0 \\-1 & \text{if } x \geq 0 \end{cases}$

Last updated date: 20th Jun 2024
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Hint:

To identify points of discontinuity, analyze the limits of $f(x)$ as $x$ approaches the boundaries where the function's definition changes. Specifically, check the limit as $x$ approaches 0 from both the left and the right.

Step-by-Step Solution:

The function is defined as:

$f(x) = \begin{cases} \dfrac{x}{|x|} & \text{if } x < 0 \\-1 & \text{if } x \geq 0 \end{cases}$

​

1. For $x < 0$ the function is given by $f(x) = \dfrac{x}{|x|}$. Since $x$ is negative in this region, $|x| = -x$. Thus, the function can be expressed as:
$f(x) = \dfrac{-x}{x} = -1$

2. Let's compute the function's value as $x$ approaches 0 from the left:

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

3. For $x \geq 0, f(x) = -1$. So, evaluating the right-hand limit:

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

4. As the limit from the left (−1) matches the limit from the right (−1) and also matches the function's value at $x = 0, f(x)$ is continuous at $x = 0$.

5. For all other x-values, the function is defined and continuous.

The function $f(x)$ is continuous everywhere.

Note:

This function is particularly interesting. For $x < 0$, the function consistently outputs −1, and for $x \geq 0$, it is defined as −1. So, its value is essentially −1 for all $x$. The utilization of the fraction $frac{x}{|x|}$ is a clever method to always obtain the result −1. Given that the function is −1 across all $x$, it remains continuous everywhere.