Q.Evaluate $\displaystyle\lim_{x\to0}\dfrac{\sin 3x}{x}$.

This has the same 'sine over its own angle' pattern as the standard limit $\lim_{x\to0}\frac{\sin x}{x}=1$, except the angle inside the sine is $3x$ while the denominator is plain $x$ — so we adjust to match the standard form exactly.

Step 1 — Match the standard form. The standard limit is $\lim_{t\to0}\dfrac{\sin t}{t}=1$, which requires the same expression in both the sine's argument and the denominator. Here the argument is $3x$ but the denominator is $x$ — so rewrite by multiplying and dividing by $3$:

Step 2 — Apply the standard limit. Let $t=3x$; as $x\to0$, $t\to0$ as well, so $\dfrac{\sin3x}{3x} = \dfrac{\sin t}{t} \to 1$.

Step 3 — Combine. $\displaystyle\lim_{x\to0}\frac{\sin3x}{x} = 3\cdot\lim_{x\to0}\frac{\sin3x}{3x} = 3\times1=3$. …

A frequent error is to write $\lim_{x\to0}\dfrac{\sin3x}{x}=1$ directly, applying the standard limit without adjusting for the mismatch between the sine's argument ($3x$) and the denominator ($x$) — the …