Q.Find $\lim_{x\to 5} f(x)$, where $f(x) = |x| - 5$.

The key idea is that $|x|$ is a polynomial-like continuous function (it is the absolute value, which is continuous everywhere), so the limit can be found by direct substitution.

Step 1: Check continuity. The function $f(x) = |x| - 5$ is continuous for all real $x$ (absolute value is continuous, subtraction preserves continuity). …

The key idea here is that polynomials and absolute value functions are continuous — meaning the limit as $x$ approaches a point is just the function’s value at that point. No factoring, no cancellation, no special tricks.

$f(x) = |x| - 5$ is a simple combination: the absolute value function (continuous everywhere) minus a constant. So the whole expression is continuous for all real $x$.

1. Check continuity at $x = 5$.

   The absolute value function $|x|$ is continuous at every real number — there’s no break, jump, or hole. Subtracting 5 shifts the graph down but doesn’t affect continuity. So $f(x)$ is continuous at $x = 5$.

2. Apply the direct substitution property for continuous functions.

   If $f$ is continuous at $x = a$, then $\lim_{x \to a} f(x) = f(a)$.

   Here $a = 5$, so:

3. Evaluate $|5|$. Since $5 \ge 0$, $|5| = 5$. Therefore: …