Sequences — Question 10

*Question 10 Let $a_n=\cos(1/n)$, where angles are measured in radians.

1. Find $\lim_{n\to\infty}a_n$ using continuity.

2. Use $0\le1-\cos x\le x^2/2$ to obtain an explicit upper bound for the error $|a_n-L|$.

3. Find a simple integer $N$ that guarantees $|a_n-L|<10^{-4}$ for every $n\ge N$.

*Question 10 – Solution

Since $1/n\to0$ and cosine is continuous at $0$, $$L=\lim_{n\to\infty}\cos(1/n)=\cos0=1.$$

Because $\cos(1/n)\le1$, the absolute error simplifies to $$|a_n-L|=1-\cos(1/n).$$ Apply $0\le1-\cos x\le x^2/2$ with $x=1/n$: $$0\le |a_n-1|\le\frac{1}{2n^2}.$$

It is enough to make the upper bound smaller than the desired tolerance: $$\frac1{2n^2}<10^{-4} \quad\Longleftrightarrow\quad n^2>5000 \quad\Longleftrightarrow\quad n>\sqrt{5000}\approx70.71.$$

Thus $N=71$ guarantees $|a_n-1|<10^{-4}$ for every $n\ge71$, because $1/(2n^2)$ decreases with $n$. This is a sufficient value obtained from an upper bound; it need not be the smallest possible $N$ for the exact cosine error.