Q.Find the values of $a$ and $b$ such that the function $f$ defined by $f(x) = \begin{cases} \dfrac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \dfrac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$ is a continuous function at $x = 4$.

Concept: Continuity at a point — set LHL $=$ RHL $=f(4)$.

The expression $\dfrac{x-4}{|x-4|}$ is the sign of $(x-4)$: it equals $-1$ for $x<4$ and $+1$ for $x>4$.

Left-hand limit ($x<4$): $\dfrac{x-4}{|x-4|}=-1$, so $\lim_{x\to 4^-}f(x)=-1+a=a-1$.

Right-hand limit ($x>4$): $\dfrac{x-4}{|x-4|}=1$, so $\lim_{x\to 4^+}f(x)=1+b=b+1$. …

The idea

$f$ is continuous at $x=4$ exactly when the left-hand limit, the right-hand limit, and $f(4)$ are all equal. The only awkward piece is $\dfrac{x-4}{|x-4|}$, which is just the sign of $x-4$.

Step-by-step

1. Left-hand limit. For $x<4$, $|x-4|=-(x-4)$, so

2. Right-hand limit. For $x>4$, $|x-4|=x-4$, so

3. Value at the point. $f(4)=a+b$.

4. Impose continuity. All three must agree: …

Method: Solving for Two Unknowns Using a Sign-Function Piece

This method applies to piecewise functions built from an expression like $\dfrac{x-a}{|x-a|}$ — which is really just the algebraic sign of $(x-a)$ — combined with two unknown constants, where continuity at $a$ must pin down both constants.

Steps

Step 1: Reduce the sign-pattern expression to $\pm 1$

Recognise that $\dfrac{x-a}{|x-a|}$ equals $-1$ for every $x<a$ and $+1$ for every $x>a$, regardless of exactly how close $x$ is to $a$. This turns each branch into a simple constant plus the unknown.

Step 2: Write the three continuity quantities in terms of the unknowns

$\lim_{x\to a^-} f(x), \qquad \lim_{x\to a^+} f(x), \qquad f(a)$ …

Common Mistakes

Mistake 1: Getting the sign of $\dfrac{x-a}{|x-a|}$ backwards on one side

Why it's wrong: it's easy to mix up which side gives $-1$ and which gives $+1$, especially when the point isn't $0$ — this flips the sign of one continuity equation and produces the wrong values for both constants. Correct approach: explicitly substitute $x<a$ and $x>a$ into $|x-a|$ separately and simplify before assigning $\pm1$.

Mistake 2: Treating "all three quantities equal" as only one equation …