Q.Show that $\displaystyle\lim_{x\to0}\frac{(1+x)^3-1}{x}$ (evaluated using the special formula) gives the same value as $\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x-a}$ evaluated at $n=3,\,a=1$ (the general formula) — confirming the two are different representations of the same result.

The special formula directly gives $n=3$; the general formula, substituted at $a=1$, must reduce to the same number.\n> [!TLDR]\n> Special form gives $3$; general form at $n=3,a=1$ gives $3(1)^2=3$ — they match.\n\nBoth routes give $3$, confirming t …

Route 1 — special ($x\to0$) form directly. By $\lim_{x\to0}\dfrac{(1+x)^n-1}{x}=n$ with $n=3$:

Route 2 — general form at $a=1$. By $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$ with $n=3,\,a=1$:

Confirming these are the same limit, not just coincidentally equal numbers: substitute $x=1+t$ in Route 2's expression (so $x\to1$ exactly when $t\to0$):

$\frac{x^3-1}{x-1} = \frac{(1+t)^3-1}{(1+t)-1} = \frac{(1+t)^3-1}{t}.$ …

An independent numeric check: for $t=0.001$, $(1+t)^3-1 = 1.003003001-1=0.003003001$, and dividing by $t=0.001$ gives approximately $3.003$ — close to $3$ a …

A common confusion is treating the two named formulae as unrelated facts to memorise separately, which doubles the memory burden and risks misapplying one where the other's pattern actually fits — recognising $a=1$ is …