Q.Evaluate $\displaystyle\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}$ using the standard algebraic limit formula.

Write $\sqrt{x}=x^{1/2}$ and $2=4^{1/2}$, matching the pattern with $n=\tfrac12,\,a=4$.\n> [!TLDR]\n> Match $n=\tfrac12$, $a=4$; apply $na^{n-1}=\tfrac12\cdot4^{-1/2}$.\n\n$\tfrac1 …

Direct substitution gives $\dfrac{\sqrt4-2}{4-4}=\dfrac{0}{0}$ — indeterminate, so the standard formula is needed (this time with a rational, non-integer exponent, exactly the general case the formula is stated for).

Rewrite $\sqrt{x}=x^{1/2}$ and note $2=\sqrt4=4^{1/2}$, so the expression matches $\dfrac{x^n-a^n}{x-a}$ with $n=\dfrac12$ and $a=4$.

Apply $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$:

Cross-check by rationalising (an independent technique): multiply numerator and denominator by the conjugate $(\sqrt{x}+2)$: …

Conjugate-rationalisation cross-check (shown above): simplifying $\frac{\sqrt x - 2}{x-4}$ to $\frac{1}{\sqrt x + 2}$ by multiplying through by the conjugate, then substitutin …

A common error with fractional $n$ is writing $4^{-1/2}$ as a negative number (confusing the negative EXPONENT with a negative VALUE) — $4^{-1/2}=\dfrac{1}{4^{1/2}}=\dfrac12$ is a …