Compute $\lim_{n\to\infty}{a_n}$ Given $a$ and $b$  such that $a \ge b &gt; 0,$ define the sequence ${a_n}$ by 
$$a_1 = a+b  , \ \ \ a_n= a_1  - \frac {ab}{a_{n-1}} , \ \ \ \ \ n \ge 2.$$

Determine the $n^{\rm th}$ term of the sequence and compute $\lim_{n\to\infty}{a_n}$

I take $l = l - \frac {ab}{l}$ after that $ l= l - \frac{ab}{l}$, $l^2 =l^2-ab$, and I get $ab = 0$.

I don't know how I can  find $\lim_{n\to\infty}{a_n}$.
The limit, assuming it exists, will satisfy
$$
L=a+b-\frac{ab}L.
$$
This has roots $a$ and $b$, so this is not enough. 

When $a=b$, we get with ease that $a_n=a(1+\frac1n)$, and so the limit is $a$. 

So now assume that $a&gt;b$. 
With a little playing with the formulas, one can guess that 
$$\tag1
a_n=\frac{a^{n+1}-b^{n+1}}{a^n-b^n}.
$$
Indeed, if $a_n$ is as above, 
\begin{align}
a_{n+1}&amp;=a+b-\frac{ab}{\frac{a^{n+1}-b^{n+1}}{a^n-b^n}}
=a+b-\frac{ab(a^n-b^n)}{a^{n+1}-b^{n+1}}\\ \ \\
&amp;=\frac{(a+b)(a^{n+1}-b^{n+1})-ba^{n+1}+ab^{n+1}}{a^{n+1}-b^{n+1}}\\ \ \\
&amp;=\frac{a^{n+2}-b^{n+2}}{a^{n+1}-b^{n+1}},\\ \ \\
\end{align}
which implies by induction that $a_n$ is as in $(1)$ for all $n$. 
Now, looking at $(1)$, and since $a&gt; b\geq0$, 
$$
a_n=a\,\frac{1-\left(\frac ba\right)^{n+1}}{1-\left(\frac ba\right)^{n}}\xrightarrow[n\to\infty]{}a.
$$
Assuming the limit exists, you would have (since $a_1 = a+b$)
$$
l = a_1 - \frac{ab}{l} \implies l^2-(a+b) l +ab =0
$$
so you get that the limit is $l=a$.