Q:

How to compute this kind of limit?

Let $x_0=a,x_1=b$

$$x_{n+1}=\Big(1-\dfrac{1}{2n}\Big)x_n+\dfrac{ x_{n-1}}{2n}, n\ge1$$

Find $\lim x_n.$
If the limit exists, I can plug limit as $l$ to get a equation of $l$, whose root will be the limit.
But in this case it becomes a identity.
I am not familiar about how to compute such a limit. Please Help. 

A:

$$x_{n+1}-x_n=-\dfrac{1}{2n}(x_n-x_{n-1})$$ 
$$\dfrac{x_{n+1}-x_n}{x_n-x_{n-1}}=-\dfrac{1}{2n}$$
 Multiplying first $n-1$ terms we can obtain
 $$\dfrac{x_n-x_{n-1}}{x_1-x_0}=\dfrac{(-1)^{n-1}}{2^{n-1}(n-1)!}$$ 
$$x_n-x_{n-1}=\dfrac{(-1)^{n-1}(b-a)}{2^{n-1}(n-1)!}$$ Now telescope!!
$$x_n-a=(b-a)\sum_{k=0}^{n-1}\dfrac{\Big(-\dfrac{1}{2}\Big)^{k}}{k!}\to (b-a)e^{-\frac{1}{2}}$$ $$\lim_{n\to\infty}x_n=a+(b-a)e^{-\frac{1}{2}}$$

