Q.A function is defined by $f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \neq 2$ and $f(2) = 5$. Examine the continuity of $f$ at $x = 2$, and if it is discontinuous, state the type of discontinuity.

The rational expression cancels for $x\neq 2$, so the limit exists; compare it with the given value $f(2)=5$.

$\dfrac{x^2-4}{x-2}=x+2$ for $x\neq2$, so $\lim_{x\to2}f(x)=4$, but $f(2)=5\neq4$ — a removable discontinuity.

Test the three conditions at $x = 2$.

Condition 1 — $f(2)$ defined? Yes, it is given as $f(2) = 5$.

Condition 2 — does the limit exist? For $x \neq 2$, factor the numerator: $\dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2} = x + 2$ (the common factor $x-2$ cancels, valid since $x \neq 2$ in the limit process). Hence $\displaystyle\lim_{x\to 2} f(x) = \lim_{x\to 2}(x+2) = 2 + 2 = 4$. The two-sided limit exists and equals $4$.

Condition 3 — does the limit equal $f(2)$? The limit is $4$ but $f(2) = 5$, so $\displaystyle\lim_{x\to 2} f(x) \neq f(2)$. Condition 3 fails.

Therefore $f$ is discontinuous at $x = 2$. Because the two-sided limit exists (it is $4$) but simply does not match $f(2)$, the discontinuity is removable — redefining $f(2) = 4$ would make $f$ continuous there.

Check (dual-solve): compute the one-sided limits separately. As $x\to 2^-$, $x+2 \to 4$; as $x\to 2^+$, $x+2 \to 4$. Both one-sided limits are finite and equal ($=4$), confirming the two-sided limit is $4$ and that the failure is only a mismatch with $f(2)=5$ — a removable discontinuity, consistent with the classification above.

The discontinuity can be removed by redefining $f(2) = 4$ (the limit value); with that single change the function $f(x)=x+2$ everywhere would be a plain polynomial, continuous at $x=2$.

A frequent mistake is to substitute $x=2$ directly into $\dfrac{x^2-4}{x-2}$, get $\dfrac{0}{0}$, and conclude the limit "does not exist" — but $\tfrac00$ is an indeterminate form, not a non-existent limit; the common factor must be cancelled first. Another is to call the discontinuity a "jump" — it is removable, because the two-sided limit genuinely exists.