<s>[INST] $\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}.$ I would like to ask why is the sequence convergent to $\dfrac{1}{2}$ when the parameter $|a|<1$ ?
At the same time, I would like to ask if there is a limit for $a<-1$, because the sequence would alternate with signs.
$$\lim_{n\to\infty} \frac{a^{2n}+1}{a^n+2}$$
 [/INST] I have tried to make the answer as simple as possible, so forgive me if it sounds a bit silly to the many experts among you....  :~)
First we simplify $\frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}$
we know that if |a|<1, then, $\lim _{x\rightarrow \infty }a^n=0$ (as when numbers smaller than one are raised to higher powers, they get smaller, eg. (0.01)^2=0.001,etc ...)
Hence,   $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{0}{2+0}+\frac{1}{2+0}=\frac{1}{2}$

1 raised to any positive power is 1, hence at a=1,  we have $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{\left(1^{\infty }+1\right)}{2+1^{\infty }}=\frac{2}{3}$

But if |a|>1, then $\lim _{x\rightarrow \infty }a^n=\infty$ (as when numbers bigger than  one are raised to higher powers, they get bigger, eg. 10^2=100,etc ...)
Hence,  $\lim _{n\rightarrow \infty }\ \ \frac{a^{2n}+1}{2+a^n}=\frac{a^{2n}}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\cdot a^n\right)}{2+a^n}+\frac{1}{2+a^n}=\frac{\left(a^n\right)}{\frac{2}{a^n}+1}+\frac{1}{2+a^n}=\frac{\infty }{\frac{2}{\infty }+1}+\frac{1}{2+\infty }=\infty $
</s>