If a sequence is defined by $x_{n+1}=\frac{4+x_n}{1+x_n}, x_1=1$ then the method for finding the limit is to recognise that $x_{n+1}$ and $x_n$ both approach the limit $L$. Then we get $L=\frac{4+L}{1+L}$. Solving for $L$ yields $L=2$ since $L\gt 0$. However this doesn't take into account that $x_1=1$, only that $L\gt0$.
Why isn't it necessary to take into account the initial value? Should different initial values create different sequences which could have different limits?
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That $L$ is the possible value, i.e. if the sequence $x_n$ converges, then it must converge to $L$. However you could give some $x_1$ s.t. the sequence being divergent.
Actually, if $x_1 \in (-4,-1)$, then $x_2 <0$, thus some $x_1$ could lead you to a negative $L$ [I have not explicitly tried, this is something popped up at the glimpse]. So actually you did consider the $x_1$: $x_1>0\implies x_n >0$ for all $n$ $\implies L \geqslant 0$.
Different initial values may create different sequences which could have different limits. If you take $x_1=-2$, then $x_n=-2$ for all $n\geq 1.$ So the limit is $-2$. So it must be consider the initial value.
You can consider the image of the function $y=\frac{4+x}{1+x}$, and it has two points of intersection $(-2,-2)$ and $(2,2)$ with the line $y=x$.