How to find a sequence by its limit?

Question

Is there any way to construct non-trivial sequence by its limit? Something like $\begin{cases} a_1=2 & \\ a_{n+1}=\dfrac1{2}\left(a_n+\dfrac2{a_n}\right) \end{cases}$
for $\sqrt2$. I'm especially interested in square roots, trigonometric functions and alike.
By non-trivial i mean that the sequence definition mustn't contain the limit itself - let's say i want to approximate it.

Answer 1

For $\sqrt{x}$, you have already almost it:

$$\begin{cases} a_1=2 & \\ a_{n+1}=\dfrac1{2}\left(a_n+\dfrac{x}{a_n}\right) \end{cases}$$

Generally speaking, given a number $x$, you can easily construct a sequence that converges to it: simply try $u_n=x+\frac1{n}$. But if you want something non trivial like this, it's absolutely not immediate in general. And usually, the way your number is defined is already a sequence or something amenable to a sequence.

For example, $\sin x$ and $\cos x$ are often defined as

$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$

$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$

These infinite series can be seen as sequences of partial series (you sum up to $m$, and you take growing and growing $m$).