Evaluate the limit of (1 - cos 2x)/(2x) as x approaches infinity.

Step-by-step solution

1. 1

   Notice the numerator never settles. As $x\to\infty$, $\cos 2x$ keeps oscillating between $-1$ and $1$ forever, so $1-\cos 2x$ has no limit. That rules out splitting the fraction or applying L'Hopital's rule — but it does not stop the whole quotient from converging, because the numerator stays bounded.

2. 2

   Bound the numerator. From $-1\le\cos 2x\le 1$ we get $0\le 1-\cos 2x\le 2$ for every real $x$. This is the key inequality: a bounded numerator over a numerator-free denominator that grows without bound.

3. 3

   Divide the inequality by the positive denominator. For $x>0$ we have $2x>0$, so dividing preserves the direction of both inequalities: $0\le\frac{1-\cos 2x}{2x}\le\frac{2}{2x}=\frac{1}{x}.$

4. 4

   Apply the squeeze theorem. Both outer bounds go to the same place: $\lim_{x\to\infty}0=0$ and $\lim_{x\to\infty}\tfrac1x=0$. A function trapped between two sequences that share a limit must share it too, so $\lim_{x\to\infty}\frac{1-\cos 2x}{2x}=0.$

5. 5

   Confirm with a spot value. At $x=1000$ the numerator can be as large as $2$, so the quotient is at most $2/2000=0.001$; at $x=10^6$ it is at most $5\times 10^{-7}$. The upper bound itself collapses to $0$, which is what forces the answer.