Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{|x - 4|}{2(x - 4)}, & x \ne 4 \\ 0, & x = 4 \end{cases}$ at $x = 4$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: Evaluate the left-hand limit.

For $x \to 4^-$, $x - 4 < 0$, so $|x - 4| = -(x - 4)$.

Then

Step 2: Evaluate the right-hand limit.

For $x \to 4^+$, $x - 4 > 0$, so $|x - 4| = x - 4$.

Then

$\lim_{x \to 4^+} \frac{|x - 4|}{2(x - 4)} = \lim_{x \to 4^+} \frac{x - 4}{2(x - 4)} = \frac{1}{2}.$ …

Concept First: Continuity at a Point

A function $f$ is continuous at $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\displaystyle \lim_{x \to a} f(x)$ exists.
3. $\displaystyle \lim_{x \to a} f(x) = f(a)$.

The tricky part is condition 2: the limit exists only when the left-hand limit and right-hand limit are equal. For piecewise functions with absolute values, the sign of the expression inside the absolute value changes at the "corner" point — here, at $x = 4$. That's why we must check both sides separately.

Step-by-step solution

1. Check the function value at $x = 4$

The definition directly gives $f(4) = 0$. So condition 1 is satisfied.

2. Compute the right-hand limit ($x \to 4^+$)

When $x > 4$, the expression $x - 4$ is positive, so $|x - 4| = x - 4$.

Thus for $x > 4$:

Therefore,

3. Compute the left-hand limit ($x \to 4^-$)

When $x < 4$, the expression $x - 4$ is negative, so $|x - 4| = -(x - 4)$.

Thus for $x < 4$:

Therefore,

4. Compare the one-sided limits

We have: …

Method: Testing Continuity of an Absolute-Value Expression via One-Sided Limits

Whenever the "$x \ne a$" branch of a piecewise function contains $|x-a|$ (or any absolute value whose argument changes sign exactly at the point being tested), a single two-sided limit computation is not valid — the expression itself is genuinely different on either side of $a$, so the one-sided limits must be found separately.

Steps

Step 1: Determine how the absolute value "splits" the domain.

Identify the point where the expression inside $|\cdot|$ changes sign — that is exactly the point being tested. Write:

Step 2: Substitute the correct branch and simplify the right-hand limit ($x \to a^+$).

Replace $|x-a|$ with $x-a$ (its value for $x>a$) inside the original expression, simplify algebraically, then take the limit.

Step 3: Substitute the correct branch and simplify the left-hand limit ($x \to a^-$). …

Common Mistakes

Mistake 1: Cancelling $|x-4|$ with $(x-4)$ as if it were simply $(x-4)$, ignoring the sign change.

Why it's wrong: $|x-4|$ equals $(x-4)$ only when $x>4$; for $x<4$ it equals $-(x-4)$. Cancelling without splitting into cases silently loses the sign, and both one-sided limits get computed as the same (wrong) value. Correct approach: always rewrite $|x-4|$ separately for $x>4$ and $x<4$ before simplifying.

Mistake 2: Treating the given value $f(4)=0$ as automatically confirming continuity, without first checking the two-sided limit exists. …