Evaluate the limit of n^2 times (1 - cos(2/n)) as n approaches infinity.

Step-by-step solution

1. 1

   Name the indeterminate form. As $n\to\infty$, $n^2\to\infty$ while $1-\cos\frac{2}{n}\to 0$, so the product is of type $\infty\cdot 0$ and must be rewritten before any limit can be read off.

2. 2

   Substitute a shrinking variable. Let $u=\frac{2}{n}$, so $n=\frac{2}{u}$ and $u\to 0^+$. Then $n^2\left(1-\cos\frac{2}{n}\right)=\frac{4}{u^2}\left(1-\cos u\right)=4\cdot\frac{1-\cos u}{u^2}$.

3. 3

   Recall the companion standard limit. $\lim_{u\to 0}\frac{1-\cos u}{u^2}=\frac{1}{2}$. It follows from the half-angle identity $1-\cos u=2\sin^2\frac{u}{2}$ together with $\lim\frac{\sin t}{t}=1$.

4. 4

   Combine the pieces. The limit is $4\cdot\frac{1}{2}=2$.

5. 5

   Cross-check with a series. Since $\cos u=1-\frac{u^2}{2}+O(u^4)$, we get $1-\cos\frac{2}{n}\approx\frac{2}{n^2}$, and multiplying by $n^2$ leaves $2$.

6. 6

   Numerical check. At $n=10^{5}$ the expression evaluates to $1.99999999993$, confirming the limit.