Q.A function is defined by $f(x) = \dfrac{x^2 - 16}{x - 4}$ for $x \neq 4$ and $f(4) = 10$. Examine the continuity of $f$ at $x = 4$ and state the type of any discontinuity.

The rational expression cancels for $x\neq4$, so the limit exists; compare it with the given $f(4)=10$.

$\dfrac{x^2-16}{x-4}=x+4$ for $x\neq4$, so $\lim_{x\to4}f(x)=8$, but $f(4)=10\neq8$ — removable discontinuity.

Test the three conditions at $x = 4$.

Condition 1. $f(4) = 10$ is defined (given).

Condition 2. For $x \neq 4$, $\dfrac{x^2 - 16}{x - 4} = \dfrac{(x-4)(x+4)}{x-4} = x + 4$. Hence $\displaystyle\lim_{x\to 4} f(x) = 4 + 4 = 8$ — the limit exists.

Condition 3. The limit $8 \neq f(4) = 10$, so condition 3 fails.

Therefore $f$ is discontinuous at $x = 4$. Since the two-sided limit exists (it is $8$) but does not match $f(4)$, the discontinuity is removable — redefining $f(4) = 8$ would restore continuity.

Check (dual-solve): both one-sided limits of $x+4$ at $x=4$ equal $8$, so the limit genuinely exists and the failure is only the mismatch with $f(4)=10$ — confirming a removable (not jump or infinite) discontinuity.

The discontinuity is removed by redefining $f(4)=8$; the function then equals $x+4$ throughout and is continuous at $x=4$.

As in Worked Example 2, do not read the raw $\tfrac00$ at $x=4$ as "no limit" — cancel the factor $x-4$ first. And do not call this a jump: a jump requires unequal finite one-sided limits, whereas here the two-sided limit exists (both sides give $8$), making it removable.