Q.Evaluate $\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}$.

Direct substitution here gives $\frac00$, an indeterminate form, so we factorise the numerator first to cancel the common factor causing the problem.

Factorising, $x^2-4=(x-2)(x+2)$, so $\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$ for $x\neq2$. Substituting $x=2$ into the simplified expression: $2+2=4$.

Step 1 — Try direct substitution. Putting $x=2$ gives $\dfrac{2^2-4}{2-2}=\dfrac{0}{0}$, an indeterminate form — direct substitution cannot be used as it stands.

Step 2 — Factorise. The numerator is a difference of squares: $x^2-4=(x-2)(x+2)$. So for every $x\neq2$,

Step 3 — Substitute into the simplified form. Since a limit only depends on values near (not at) $x=2$, we may use the simplified expression: $\lim_{x\to2}(x+2) = 2+2=4$.

Check (independent verification). Numerically at $x=1.999$: $\dfrac{1.999^2-4}{1.999-2} = \dfrac{3.996001-4}{-0.001} = \dfrac{-0.003999}{-0.001}=3.999$, very close to $4$, confirming the algebraic result.

Alternative: substitute $x=2+h$ and let $h\to0$: $\dfrac{(2+h)^2-4}{h} = \dfrac{4+4h+h^2-4}{h} = \dfrac{4h+h^2}{h} = 4+h \to 4$ as $h\to0$ — the same answer by a substitution approach instead of factorisation.

Concluding the limit 'does not exist' simply because direct substitution gives $\frac00$ is a serious misunderstanding — $\frac00$ signals that substitution is the wrong tool here, not that the limit fails to exist. Also, students sometimes cancel incorrectly, e.g. cancelling an $x$ that isn't actually a common factor of both numerator and denominator.