Which of the following limits does not exist?
A) $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left| x \right|}}{x}$
B) $\mathop {\lim }\limits_{x \to 0} \{ x + \left| x \right|\} $
C) $\mathop {\lim }\limits_{x \to 0} \left| x \right|$
D) $\mathop {\lim }\limits_{x \to 0} \{ x - \left| x \right|\} $

Hint: To solve this question, we need to know the basic definition of limit.
When we write: $\mathop {\lim }\limits_{x \to p} f(x) = L$ , it explains how a real function $f(x)$ reaches a value $L$ when $x$ tends to $p$ .
When we say that a limit exists, we mean that:
$\mathop {\lim }\limits_{x \to {p^ + }} f(x) = \mathop {\lim }\limits_{x \to {p^ - }} f(x) = \mathop {\lim }\limits_{x \to p} f(x) = L$
The above equation tells us that: a real function $f(x)$ reaches a value $L$ when $x$ tends to $p$ from both left side and right side. When this happens, we can say that a function is continuous.
A limit does not exist at a certain value of $x$ if the function is discontinuous at that value of $x$ .
Let’s use the above concepts and solve every option.

Complete step by step solution:
A function is said to be continuous at $x$ if and only if the limit exists at that certain value of $x$ . When we say a limit exists, we mean that:
$\mathop {\lim }\limits_{x \to {p^ + }} f(x) = \mathop {\lim }\limits_{x \to {p^ - }} f(x) = L$
The first limit is known as right-handed limit. The second limit is known as left-handed limit. When both of them are equal to the function value at that $x$ , we say it is continuous. For a function at a certain value of $x$, the limit doesn’t exist if it is discontinuous.
$\left| x \right|$ is modulus function which returns the absolute value of the function, it is always greater than or equal to zero.
Now let’s solve this question by calculating the limit for all the options.
Option A: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left| x \right|}}{x}$
$\dfrac{{\left| x \right|}}{x}$ is also known as signum function.
The right-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left| x \right|}}{x}$
When $x \to {0^ + }$ , $\left| x \right| = x$
Therefore, the limit value becomes: $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left| x \right|}}{x} = 1$
The left-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\left| x \right|}}{x}$
When $x \to {0^ - }$ , $\left| x \right| = - x$
Therefore, the limit value becomes: $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\left| x \right|}}{x} = - 1$
Since, the left-handed limit is not equal to the right-handed limit, the limit does not exist.
Option B: $\mathop {\lim }\limits_{x \to 0} \{ x + \left| x \right|\} $
When we say \[\left| x \right|\]
when $x < 0$ , $\left| x \right| = - x$
when $x > 0$ , $\left| x \right| = x$
the left-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ - }} \{ x + \left| x \right|\} $
After opening the modulus value, we get:
$\mathop {\lim }\limits_{x \to {0^ - }} \{ x + \left| x \right|\} = \mathop {\lim }\limits_{x \to {0^ - }} \{ x + ( - x)\} = \mathop {\lim }\limits_{x \to {0^ - }} 0 = 0$
The right-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ + }} \{ x + \left| x \right|\} $
After opening the modulus, we get:
$\mathop {\lim }\limits_{x \to 0 + } \{ x + \left| x \right|\} = \mathop {\lim }\limits_{x \to 0 + } \{ x + x\} = \mathop {\lim }\limits_{x \to 0 + } 2x$
After substituting the value of $x$ , we get: $\mathop {\lim }\limits_{x \to 0 + } 2x = 0$
Since, the left-handed and right-handed limits are equal, the limit exists.
Option C: $\mathop {\lim }\limits_{x \to 0} \left| x \right|$
The left-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ - }} \left| x \right|$
Opening the modulus, we get:
$\mathop {\lim }\limits_{x \to {0^ - }} \left| x \right| = \mathop {\lim }\limits_{x \to {0^ - }} ( - x) = 0$
The right-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ + }} \left| x \right|$
Opening the modulus, we get:
$\mathop {\lim }\limits_{x \to {0^ + }} \left| x \right| = \mathop {\lim }\limits_{x \to {0^ + }} x = 0$
Since, the right-handed and left-handed limits are equal, the limit exists.
Option D: $\mathop {\lim }\limits_{x \to 0} \{ x - \left| x \right|\} $
Using the same simplification technique we used in option B, we get:
The left-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ - }} \{ x - \left| x \right|\} $
After further simplification, we get:
$\mathop {\lim }\limits_{x \to {0^ - }} \{ x - \left| x \right|\} = \mathop {\lim }\limits_{x \to {0^ - }} \{ x - ( - x)\} = \mathop {\lim }\limits_{x \to {0^ - }} 2x = 0$
The right-handed limit will be: $\mathop {\lim }\limits_{x \to {0^ + }} \{ x - \left| x \right|\} $
From further simplification, we get:
\[\mathop {\lim }\limits_{x \to {0^ + }} \{ x - \left| x \right|\} = \mathop {\lim }\limits_{x \to {0^ + }} \{ x - (x)\} = \mathop {\lim }\limits_{x \to {0^ + }} 0 = 0\]
Since, the right-handed limit and left-handed limit are equal, the limit exists.

Therefore, the correct option is A

Note: The terminology involving limits is:
A limit doesn’t exist if it is equal to an unbounded value or if the left-handed and right-handed limits are not equal.
A limit is said to vanish if it is equal to zero.
If there is a fraction, when we are finding the limit by direct substitution, we may come across undefined forms such as $\dfrac{0}{0},\dfrac{\infty }{\infty }$ , we can use L Hospital rule which is:
$\mathop {\lim }\limits_{x \to a} = \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} = \dfrac{{f'(x)}}{{g'(x)}}$