Question

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by \[ f(x) = \begin{cases} \frac{|x-4|}{x-4}, & x \neq 4 0, & x = 4 \end{cases} \] then $\lim_{x \to 4} f(x)$ is

• A. 1
• B. $e$
• C. $-1$
• D. does not exist.

Hint:

The function $f(x) = \frac{|x-a|}{x-a} = \text{sgn}(x-a)$ is the signum function shifted to $a$. Its limit at $x=a$ never exists because $\text{LHL} = -1 \neq 1 = \text{RHL}$.

Answer 1

The Correct Option is D

Step 1: Concept
For a limit $\lim_{x \to a} f(x)$ to exist, both the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) at $x = a$ must exist independently and be strictly equal to each other: \[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L \]

Step 2: Key Formulas and Approach
Recall the definition of the absolute value function: \[ |x - 4| = \begin{cases} x - 4, & \text{if } x > 4 -(x - 4), & \text{if } x < 4 \end{cases} \] We evaluate LHL ($\lim_{x \to 4^-} f(x)$) and RHL ($\lim_{x \to 4^+} f(x)$).

Step 3: Step-by-step Explanation

• Right-Hand Limit (RHL): For $x > 4$, we have $|x - 4| = x - 4$. \[ \text{RHL} = \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \frac{x - 4}{x - 4} = \lim_{x \to 4^+} 1 = 1 \]
• Left-Hand Limit (LHL): For $x < 4$, we have $|x - 4| = -(x - 4)$. \[ \text{LHL} = \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \frac{-(x - 4)}{x - 4} = \lim_{x \to 4^-} (-1) = -1 \]
• Comparing LHL and RHL: \[ \text{RHL} = 1 \quad \text{and} \quad \text{LHL} = -1 \] Since $\text{LHL} \neq \text{RHL}$ ($-1 \neq 1$), the overall limit $\lim_{x \to 4} f(x)$ does not exist.

Step 4: Final Answer
Because the left-hand limit and right-hand limit are unequal, $\lim_{x \to 4} f(x)$ does not exist. Thus, Option (D) is correct.