# What does this sequence converge to?

Suppose that a sequence $\{a_n\}$ converges to $\pi$. Then the sequence $\{\cos(a_n)\}$ _____.

The answer is converges to $-1$ but why is that case? I thought since $\cos(\pi) = 0$, the sequence should converge to 0?

Thanks! :)

-
Sorry for the poor formatting, anyone knows how to add a subscript n? – inggumnator Aug 10 '14 at 5:31
You might have to look the Mathjax tutorial, but where is the link...? Besides, does the answer really $1/e$? $\cos$ is continuous on $\Bbb R$. – The Great Seo Aug 10 '14 at 5:34
Omg sorry I copied wrongly! It converges to -1 actually. – inggumnator Aug 10 '14 at 5:37
As The Great Seo mentioned, $\cos \pi = -1$ not $0$. There lies your problem. – JimmyK4542 Aug 10 '14 at 5:38

$\cos$ is continuous on $\Bbb R$, so $$\lim_{n\to\infty} \cos(a_n)=\cos(\lim_{n\to\infty} a_n)=\color{red}{\cos(\pi)=-1}.$$

-

This is due to the continuity of $\cos(x)$. Generally a function $f$ is continuous if $$\lim_{x \to a} f(x) = f(a)$$ which in qualitative terms says if I move $x$ closer and closer to $a$ then $f(x)$ will get closer and closer to $f(a)$ (this getting closer and closer occurs at about the same rate).

Now in some simplistic terms we can consider a sequence $\{a_n\}$ to converge to a point $\pi$ to mean that as $n$ increases unboundedly, the terms of the sequence get closer and closer to $\pi$.

Intuitively, then, we can make the jump that since $a_n$ gets closer and closer to $\pi$ we can say that $$\lim_{n \to \infty} f(a_n) = f\left( \lim_{n \to \infty} a_n \right)$$

Of course this can all be proven rigorously (which I'll gladly show you how if you want) but I hope this provides some intuition.

Also note that $\cos \pi = -1$ not $0$ ;D

-
Thanks for the comprehensive answer! :) – inggumnator Aug 11 '14 at 3:37