Question:medium

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

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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 $\text{LHL} = -1 \neq 1 = \text{RHL}$.
Updated On: Jul 29, 2026
  • 1
  • $e$
  • $-1$
  • does not exist.
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The Correct Option is D

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