Question

Let the function $f(x)$ be defined as \[ f(x)=\frac{x-|x|}{x} \] then:

• A. the function is continuous everywhere
• B. the function is not continuous
• C. the function is continuous when $x<0$
• D. the function is continuous for all $x$ except zero

Hint:

Whenever modulus functions appear, always split the problem into two cases: \[ x>0 \quad \text{and} \quad x<0. \] This immediately reveals continuity and differentiability behavior.

Answer 1

The Correct Option is D

Concept: The modulus function changes definition depending on the sign of $x$: \[ |x|= \begin{cases} x, & x\ge0 -x, & x<0 \end{cases} \] Therefore, we first express the given function in piecewise form.

Step 1: Evaluating the function for $\mathbf{x>0}$.
For $x>0$, \[ |x|=x \] Substitute into the function: \[ f(x)=\frac{x-x}{x} \] \[ f(x)=0 \] Thus, for all positive values of $x$, \[ f(x)=0 \] which is continuous.

Step 2: Evaluating the function for $\mathbf{x<0}$.
For $x<0$, \[ |x|=-x \] Substituting: \[ f(x)=\frac{x-(-x)}{x} \] \[ f(x)=\frac{2x}{x} \] \[ f(x)=2 \] Thus, for all negative values of $x$, \[ f(x)=2 \] which is also continuous.

Step 3: Checking continuity at $\mathbf{x=0}$.
At $x=0$, the denominator becomes zero. Therefore, the function is undefined at $x=0$. Also, \[ \lim_{x\to0^+}f(x)=0 \] and \[ \lim_{x\to0^-}f(x)=2 \] Since the left-hand limit and right-hand limit are not equal, the function is discontinuous at $x=0$. Hence, the function is continuous everywhere except at zero. \[ \boxed{ \text{The function is continuous for all }x\neq0 } \]