Limit of sin 2x / x as x Approaches 0 Is 2

Force the argument to match the denominator

The special limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$ demands that the sine's argument and the denominator be the same expression. Here they are $2x$ and $x$, so multiply by $\frac{2}{2}$, which changes nothing, and park one 2 outside.

Now the fraction is the special limit in the variable $u = 2x$, and $u \to 0$ whenever $x \to 0$, so it tends to 1. The constant rides along untouched.

The double-angle identity lands in the same place: $\sin 2x = 2\sin x\cos x$, so the quotient is $2 \cdot \frac{\sin x}{x} \cdot \cos x$, and the three factors give $2 \cdot 1 \cdot 1$.

Why substitution fails and why the answer is not 1

Substituting gives $\frac{\sin 0}{0} = \frac{0}{0}$, indeterminate. The more expensive error comes next: a student who has memorised the sine limit sees a sine over an $x$ and writes 1 by pattern match, without checking that the arguments agree.

They do not agree, and the mismatch is the entire answer. For small $x$ the numerator behaves like $2x$ while the denominator is $x$, so the quotient sits near 2. Doubling the angle doubles the numerator, and nothing in the expression cancels that.

Common mistakes

• Answering 1. The special limit applies only after the rewrite; $\frac{\sin 2x}{x}$ is not yet in the form $\frac{\sin u}{u}$.
• Multiplying by 2 without dividing by 2. The step $\frac{\sin 2x}{x} = 2 \cdot \frac{\sin 2x}{2x}$ keeps the expression equal to the original; changing the denominator to $2x$ on its own changes the function.
• Cancelling the 2s or pulling the 2 out of the sine. $\sin 2x$ is not $2\sin x$, and no cancellation with the denominator is available.
• Rebuilding the rewrite from scratch every time instead of using $\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}$.
• Dropping the chain rule inside L'Hopital. The rule gives $\frac{2\cos 2x}{1} \to 2$, and the 2 on top is exactly the coefficient the algebraic rewrite exposes; differentiating $\sin 2x$ as $\cos 2x$ loses it.