The given function is $\mathit{f}\left(\mathit{x}\right)=|\mathit{x}|-|\mathit{x}+1|$.
The two functions, $\mathit{g}$ and $\mathit{h}$, are defined as
$\mathrm{g}\left(\mathrm{x}\right)=|\mathrm{x}|$ and $\mathrm{h}\left(\mathrm{x}\right)=|\mathrm{x}+1|$
Then, $\mathrm{f}=\mathrm{g}-\mathrm{h}$
The continuity of $\mathit{g}$ and $\mathit{h}$ is examined first.
$\mathrm{g}\left(\mathrm{x}\right)=|\mathrm{x}|$ can be written as

Clearly, $\mathit{g}$ is defined for all real numbers.
Let $\mathrm{c}$ be a real number.
Case I
If , then $\mathit{g}\left(\mathit{c}\right)=-\mathit{c}$ and $\underset{\overline{\mathit{n}}\to \mathit{c}}{\lim }\mathit{g}\left(\mathit{x}\right)=\underset{\mathit{x}\to \mathit{c}}{\lim }\left(-\mathit{x}\right)=-\mathit{c}$
$\therefore \underset{\mathit{x}\to \mathit{c}}{\lim }\mathit{g}\left(\mathit{x}\right)=\mathit{g}\left(\mathit{c}\right)$
Therefore, $\mathrm{g}$ is a continuous at all points $\mathrm{x}$, such that Case II
If $\mathit{c}>$, then $\mathit{g}\left(\mathit{c}\right)=\mathit{c}$ and $\underset{\mathit{x}\to \mathit{c}}{\lim }\mathit{g}\left(\mathit{x}\right)=\underset{\mathit{x}\to \mathit{c}}{\lim }\mathit{x}=\mathit{c}$
$\therefore \underset{\mathit{x}\to \mathit{c}}{\lim }\mathit{g}\left(\mathit{x}\right)=\mathit{g}\left(\mathit{c}\right)$
Therefore, $\mathit{g}$ is continuous at all points $\mathrm{x}$, such that $\mathrm{x}>0$
Case III
If $\mathrm{c}=\mathrm{o}$, then $\mathrm{g}\left(\mathrm{c}\right)=\mathrm{g}\left(\mathrm{o}\right)=\mathrm{o}$
$\underset{\mathit{x}\to 0}{\lim }\mathit{g}\left(\mathit{x}\right)=\underset{\mathit{x}\to 0}{\lim }\left(-\mathit{x}\right)=0$
$\underset{\mathit{x}\to 0}{\lim }\mathit{g}\left(\mathit{x}\right)=\underset{\mathit{x}\to 0}{\lim }\left(\mathit{x}\right)=0$
$\therefore \underset{\mathit{x}\to 0}{\lim }\mathit{g}\left(\mathit{x}\right)=\underset{\mathit{x}\to 0}{\lim }\left(\mathit{x}\right)=\mathit{g}\left(0\right)$
Therefore, $\mathrm{g}$ is continuous at $\mathrm{x}=\mathrm{o}$
From the above three observation, it can be concluded that $\mathrm{g}$ is continuous at all points.
$\mathrm{h}\left(\mathrm{x}\right)=|\mathrm{x}+1|$ can be written as


Clearly, $\mathrm{h}$ is defined for every real number.
Let c be a real number.
Case I :
If , then $\mathit{h}\left(\mathit{c}\right)=-\left(\mathit{c}+1\right)$ and $\underset{\overline{\mathit{n}}\to \mathit{c}}{\lim }\mathit{h}\left(\mathit{x}\right)=\underset{\mathit{x}\to \mathit{c}}{\lim }\left[-\left(\mathit{x}+1\right)\right]=-\left(\mathit{c}+1\right)$
$\therefore \underset{\mathrm{h}\to \mathrm{c}}{\lim }\mathrm{h}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{c}\right)$
Therefore, $\mathrm{h}$ is continuous at all points $\mathrm{x}$, such that
Case II:
If $\mathit{c}>-1$, then $\mathit{h}\left(\mathit{c}\right)=\mathit{c}+1$ and $\underset{\mathit{x}\to \mathit{c}}{\lim }\mathit{h}\left(\mathit{x}\right)=\underset{\mathit{x}\to \mathit{c}}{\lim }\left(\mathit{x}+1\right)=\mathit{c}+1$
$\therefore \underset{\mathrm{n}\to \mathrm{c}}{\lim }\mathrm{h}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{c}\right)$
Therefore, $\mathit{h}$ is continuous at all points $\mathit{x}$ such that $\mathit{x}>-1$.
Case III
If $\mathit{c}=-1$, then $\mathit{h}\left(\mathit{c}\right)=\mathit{h}\left(-1\right)=-1+1=0$
$\underset{\mathit{x}\to -1}{\lim }\mathit{h}\left(\mathit{x}\right)=\underset{\mathit{x}\to -1}{\lim }\left[-\left(\mathit{x}+1\right)\right]=-\left(-1+1\right)=0$
$\underset{\mathit{x}\to -1}{\lim }\mathit{h}\left(\mathit{x}\right)=\underset{\mathit{x}\to -1}{\lim }\left(\mathit{x}+1\right)=\left(-1+1\right)=0$
$\therefore \underset{\mathit{x}\to -1}{\lim }\mathit{h}\left(\mathit{x}\right)=\underset{\mathit{x}\to -1}{\lim }\mathit{h}\left(\mathit{x}\right)=\mathit{h}\left(-1\right)$
Therefore, $\mathrm{h}$ is continuous at $\mathrm{x}=1$
From the above three observations, it can be concluded that $\mathit{h}$ is continuous at all points of the real line.
$\mathrm{g}$ and $\mathrm{h}$ are continuous functions. Therefore, $\mathrm{f}=\mathrm{g}\mathrm{h}$ is also a continuous function.
Therefore, f has no point of discontinuity.