If $a_{n+1}=\frac{10}{a_n}-3$ and $a_1=10$, find $\lim_{n \to \infty} a_n$ 
Let $a_{n+1}=\dfrac{10}{a_n}-3$, $a_1=10$ then find the limit $\lim\limits_{n \to \infty} a_n$

My Try :
$$a_2=-2 \ \ ,a_3=-8 \ \, a_4=-4.25 \ \ a_n <0$$
thus visthe monotone convergence theorem $$\lim\limits_{n \to \infty}=l$$
so: $$l=\dfrac{10}{l}-3 \to l^2+3l=10 \to l=2 , -5 $$
it is right ?
 A: There is no monotone convergency with this sequence 
$$a_1=10, a_2=-2, a_3=-8, a_4=-4.25, a_5=-5.35295, a_6=-4.86813$$
is not monotone. It oscillates, so different techniques are required.
Let's look at the function $f(x)=\frac{10}{x}-3$, s.t $a_{n+1}=f(a_n)$ and $f'(x)=-\frac{10}{x^2}$

Preliminary. You have spotted a fixed point $x_0=-5$ which is also an attracting fixed point since $\left|f'(x_0)\right|=\frac{2}{5}<1$. So, there is a vicinity of $x_0$ such that if $a_n$ falls into it from some $n$ onwards, then the sequence will be "attracted" to $x_0$. This is good but not enough.

Technical proof, using Banach fixed-point theorem (BFPT). We see that starting from $n=3$, $a_n\in [-10,-4]$, this is true because
$$\color{red}{x \in [-10,-4]} \Rightarrow
-10 \leq x \leq -4 \Rightarrow \\
 10 \geq -x \geq 4 \Rightarrow  
-\frac{1}{4} \leq \frac{1}{x} \leq -\frac{1}{10} 
\Rightarrow \\
-10<-\frac{11}{2} \leq \frac{10}{x} -3 \leq -4 \Rightarrow \\
\color{red}{f(x) \in [-10,-4]}$$
And applying MVT
$$\forall x<y \in [-10,-4], \exists \xi \in (x,y) : \left|f(x)-f(y)\right|=|f'(\xi )||x-y|\leq\left|\frac{10}{16}\right||x-y|$$
Because $\left|\frac{10}{16}\right|<1$, according to BFPT, there is a limit of the sequence $a_n$ on $[-10,-4]$ (obviously unique), so it's $-5$. More theory here.
A: 
When you have a sequence of the form $a_{n+1}=f(a_n)$ that apparently does not lead to a closed formula for $a_n$, then you have to study the function $f(x)$.
When you graph the curve $y=f(x)=\dfrac{10}x-3$ in blue and $y=x$ in red, you notice there are two intersection points.
These are called fixed points of $f$ since $f(x)=x$. Once solved this gives $x=2$ or $x=-5$.
If the sequence would converge to $\ell$, the continuity of $a_{n+1}=f(a_n)$ will lead to $f(\ell)=\ell$ so $\ell$ will be one of the two fixed points.
On the graph we can see that $2$ is a repulsive point, and $-5$ an attractive point.

Since we are not required to do the full study for all initial seeds fo the sequence, but only for $a_1=10$, we will focus on showing it converges to $-5$.
We can see that the convergence is not a staircase (monotonic convergence) but a spiral. This means we have to show that $a_{2n}$ and $a_{2n+1}$ are both monotonic but of opposite direction.
To prove this we have to:


*

*study the sign of $a_{n+2}-a_n$, this is equivalent of studying the sign of $f(f(x))-x$.

*show that $-5$ is squeezed between $a_n$ and $a_{n+1}$, this is equivalent of studying the sign of $f(x)+5$.



First notice that $x<0\implies f(x)<0$ so as soon as $a_{n_0}<0$ then all subsequent $a_n$ with $n\ge n_0$ are also negative. 
Since $a_2<0$ we will select $n_0=2$.
$f(x)+5=\dfrac {10}x-3+5=\dfrac{2(x+5)}x\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$
So if $a_n<-5$ then $a_{n+1}>-5$ and vice-versa and $-5$ is squeezed between $a_n$ and $a_{n+1}$ for $n\ge 2$
$f(f(x))-x=\dfrac{10}{\frac{10}{x-3}}-3-x=\dfrac{3(x+5)(x-2)}{10-3x}\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$
So $a_{n+2}>a_{n}$ for $a_n<-5$ and $a_{n+2}<a_n$ for $a_n>-5$.
Since $a_2>-5$ then $\begin{cases}a_{2n}>-5 & a_{2n}\searrow\\a_{2n+1}<-5 & a_{2n+1}\nearrow\end{cases}$
Now we can apply monotonicity theorem and $a_{2n}\to -5$ and $a_{2n+1}\to -5$.
This means that $a_n\to -5$.
A: Note that once $x$ becomes negative, it remains negative for the rest of time; so we might as well start with $a_3 = -8$ and continue under the fact that $a_n < 0$.
Now, consider $$a_{n+2} = \frac{19 a_n-30}{10-3a_n}$$
Note that this sequence is decreasing in the even terms, and increasing in the odd terms, given the two starting values of $-8$ and $-\frac{17}{4}$: this is because for $x < -5$, we have $-6 + \frac{30+x}{10-3x} < -5$, so inductively the odd-index terms are all less than $-5$; and similarly the even-index terms are all greater than $-5$.
Hence the even terms are decreasing and bounded below; and the odd terms are increasing and bounded above.
Since the two subsequences do converge, it's now legal to do your trick and show that they converge to $-5$ by solving $l = \frac{19l - 30}{10-3l}$.

Note that one can find explicit closed forms for the even and the odd sequences, using generating functions. But that is massive overkill here.
