Question

Examine the continuity at $x = 0$: $f\left( x \right) = 1 + \dfrac{{\left| x \right|}}{x}$ for $x \ne 0$ and $f\left( 0 \right) = 1$.

Answer 1

Hint : Here, in the question, we have been given a function in the form of variable $x$. And we are asked to examine the continuity of the given function at all the points except $x = 0$. We will check the continuity of the function at $x < 0$ and $x > 0$ separately and reach a desired conclusion.

Complete step by step solution:
Given $f\left( x \right) = 1 + \dfrac{{\left| x \right|}}{x}$
And $f\left( 0 \right) = 1$
Let us first understand the meaning of $\left| x \right|$
Let $g\left( x \right) = \left| x \right|$
Then, $g\left( x \right)$ is defined as:

{\;x\;if\;x \geqslant 0} \\\ { - x\;if\;x < 0} \end{array}} \right.$$ According to this definition, if the value of $$x$$ is greater than or equal to $$0$$, then the value of modulus of $$x$$ remains the same as the value of $$x$$ but if the value of $$x$$ is less than zero, then the value of modulus of $$x$$ becomes the negative of $$x$$. Now, we have $$f\left( x \right) = 1 + \dfrac{{\left| x \right|}}{x}$$. Therefore The given function is defined at the given point $$x = 0$$ and its value is $$1$$. Now we will calculate the left hand limit and the right hand limit for the function at given point $$x = 0$$. The left hand limit of $$f$$ at $$0$$ is:

\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} 1 + \dfrac{{\left( { - x} \right)}}{x} \\
\\
\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 0 ;

The right hand limit of $$f$$ at $$0$$ is:

\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} 1 + \dfrac{x}{x} \\
\\
\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 2 ;

Since, Left hand limit of the function is not equal to the right hand limit, the function is discontinuous at $$x = 0$$. **Note** : Whenever we face such types of questions, we just have to check the left hand limits and right hand limit of the given function especially in case when continuity is being checked at $$x = 0$$. If the left hand limit, right hand limit and the value of function at $$x = 0$$ are all equal then the function is continuous for all values of $$x$$. In case, the left hand limit is equal to the right hand limit but not equal to the value of the function at a given point, then the function will be discontinuous at that point.