The answer is not 1

Write out $\frac{x}{1-x-x^2}$ as a power series and manipulate

$\begin{aligned} \frac x{1-x-x^2}&=\sum^\infty_{n=1}a_nx^n\\ x&=(1-x-x^2)\sum^\infty_{n=1}a_nx^n\\ x&=\sum^\infty_{n=1}a_nx^n-\sum^\infty_{n=1}a_nx^{n+1}-\sum^\infty_{n=1}a_nx^{n+2}\\ x&=a_1x+a_2x^2-a_1x^2+\sum^\infty_{n=3}(a_n-a_{n-1}-a_{n-2})x^n\\ x&=a_1x+(a_2-a_1)x^2+\sum^\infty_{n=3}(a_n-a_{n-1}-a_{n-2})x^n\\ \end{aligned}$

Now relate the coefficients. this gives $a_1=1$ and $a_2-a_1=0\implies a_2=a_1=1$ . Then the other coefficients have to equal zero as well meaning we have the recurrence relation $a_n-a_{n-1}-a_{n-2}=0\implies a_n=a_{n-1}+a_{n-2}$ .

Now because $a_1=a_2=1$ and we have an addition relation it means that $a_{n}=F_n$ where $F_n$ are the Fibonacci numbers.

Therefore $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \frac{F_{n+1}}{F_{n}}=\phi\approx\boxed{1.618}$ The last one is a well known limit.

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Edit: To expand on deriving the limit

$\begin{aligned} L&=\lim_{n\to\infty} \frac{F_{n+1}}{F_n} \\ L&=\lim_{n\to\infty} \frac{F_n+F_{n-1}}{F_n} \\ L&=1+\lim_{n\to\infty} \frac{F_{n-1}}{F_n} \\ L&=1+\frac1{\displaystyle\lim_{n\to\infty} \frac{F_{n}}{F_{n+1}}} \\ L&=1+\frac1L\\ \end{aligned}$

The last equation can be solved easily. It is the definition of the golden ratio and can be turned into a quadratic to yield it also. But be careful about introducing a new solution when turning it into a quadratic.

We can show that $\frac{x}{1-x-x^2}$ is the generating function of the Fibonacci number by long division method. It rather difficult to show here.

$\dfrac{x}{1-x-x^2}=\dfrac{x(1-x-x^2)+x^2+x^3}{1-x-x^2}=x+ \dfrac{x^2+x^3}{1-x-x^2}$

$=x+\dfrac{x^2(1-x-x^2)+2x^3+x^4}{1-x-x^2}=x+ x^2 + \dfrac {2x^3+x^4} {1-x-x^2}$

$=x+x^2 +\dfrac{2x^3(1-x-x^2)+3x^4+2x^5}{1-x-x^2}=x+ x^2 + 2x^3 + \dfrac {3x^4+2x^5} {1-x-x^2}$

$=x+x^2 + 2x^3 + \dfrac{3x^4(1-x-x^2)+5x^5+3x^6}{1-x-x^2}$

$=x+ x^2 + 2x^3 + 3x^4 \dfrac {5x^5+3x^6} {1-x-x^2}$

$=x+x^2 + 2x^3 + 3 x^4+ \dfrac{5x^5(1-x-x^2)+8x^6+5x^7}{1-x-x^2}$

$= x+ x^2 + 2x^3 + 3x^4 + 5x^5 + \dfrac {8x^6+5x^7} {1-x-x^2}$

$= x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + ...$

The power-series expansion of $\frac{x}{1-x-x^2}$ is valid for all (and only) $|x|<\alpha$ , where $\alpha$ is the positive zero of the denominator, i.e. $1-\alpha-\alpha^2=0$ , i.e., $\alpha=\phi$ , the golden ratio.

Now from the ratio-test, we know it is sufficient that $\lim_{n\to \infty}\frac{|a_{n+1}||y|}{|a_n|}<1$ for the series to converge at $x=y$ . Since $\alpha$ is the radius of convergence, at $x=\alpha$ the above ratio should be one. Hence $\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|}=\frac{1}{\phi}$ . Also, from the series expansion it is evident that all the coefficients are non-negative. Thus $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\frac{1}{\phi}$ .

Doing a long division, we quickly discover that, whenever the right hand side converges (i.e. when |x|<1): $\frac{x}{1-x-x^2}=x+x^2+2x^3+3x^4+5x^5+...+F_nx^n+...$ Where $F_n$ is the nth fibonacci number (in addition, we define $F_0=0$ ). This implies that the nth-order derivative of this expression has one constant term: $n! F_n$ . So the nth order derivatife evaluated at 0 : $f^{(n)}(0)=n! F_n$

The maclaurin series (or Taylor expansion in 0) of f is

$f(x)=\sum_{n=0}^{\infty}{\frac{f^{(n)}(0)x^n}{n!}}= \sum_{n=0}^{\infty}{F_n x^n}$

and hence the requested expression equals $φ=\frac{\sqrt{5}+1}{2}=1.61803...$

It is well-known that $\dfrac{x}{1-x-x^2}$ is the generating function of the fibonacci numbers. Thus, $\lim\limits_{n\to \infty} \dfrac{a_{n+1}}{a_n}=\varphi=\boxed{1.618}$

If we need to prove that $\dfrac{x}{1-x-x^2}$ is the generating function of the fibonacci numbers, then have fun guys.