Q.Evaluate $\lim_{x\to 0} f(x)$, where $f(x) = \begin{cases} \dfrac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

The limit of a piecewise function at a point requires checking whether the left-hand and right-hand limits agree.

For $x \neq 0$, we have $f(x) = \frac{|x|}{x}$. The absolute value behaves differently on either side of zero:

Right-hand limit ($x \to 0^+$): When $x > 0$, $|x| = x$, so

Left-hand limit ($x \to 0^-$): When $x < 0$, $|x| = -x$, so

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{-x}{x} = -1$ …

The absolute value $|x|$ behaves differently on either side of zero: it equals $x$ when $x > 0$ and $-x$ when $x < 0$. This split personality means $f(x) = |x|/x$ will take different constant values depending on which side we approach from. A limit exists at a point only when both one-sided limits exist and agree.

The value $f(0) = 0$ given in the piecewise definition is irrelevant for the limit—limits care only about nearby behavior, not the point itself.

Step-by-step evaluation

1. Right-hand limit (approaching from positive side):

   For $x > 0$, we have $|x| = x$, so

As $x \to 0^+$, every value is $1$, hence

2. Left-hand limit (approaching from negative side):

   For $x < 0$, we have $|x| = -x$, so

As $x \to 0^-$, every value is $-1$, hence

3. Comparing the one-sided limits:

   Since $\lim_{x \to 0^+} f(x) = 1 \neq -1 = \lim_{x \to 0^-} f(x)$, the two-sided limit does not exist. …