Step 1: Evaluating left-hand limit (LHL).
For \( x<0 \), we have \( |x| = -x \), so
\[
f(x) = \frac{x + (-x)}{x} = \frac{0}{x} = 0
\]
Thus,
\[
\lim_{x \to 0^-} f(x) = 0
\]
Step 2: Evaluating right-hand limit (RHL).
For \( x>0 \), we have \( |x| = x \), so
\[
f(x) = \frac{x + x}{x} = \frac{2x}{x} = 2
\]
Thus,
\[
\lim_{x \to 0^+} f(x) = 2
\]
Step 3: Checking if the limit exists.
Since \( \lim_{x \to 0^-} f(x) = 0 \) and \( \lim_{x \to 0^+} f(x) = 2 \),
\[
\lim_{x \to 0} f(x) { does not exist}
\]
Thus, the correct answer is (C).