Vtyjkyuk

Given $left a_n:::n=1,:2,:3,:cdotsright $ is an infinite
sequence in $mathbbR$, and every term is positive. How to prove
that the set

beginequation
left frac2+a_nsqrt2+a_n^2:::n=1,:2,:3,:cdotsright
endequation
has a limit point?

Bolzano-Weierstrass theorem says that every bounded
infinite subset of $mathbbR$ has a limit point. But how to prove
it is bounded? I tried this
beginequation
left|frac2+a_nsqrt2+a_n^2right|leqleft|frac2+a_nsqrta_n^2right|=left|frac2+a_na_nright|
endequation
but does not seem work.

Or these is another way to show it has a limit point, without using Bolzano-Weierstrass theorem

Hint: You can let intuition guide you: When $a_n$ is very small, it is negligible compared to $2$, so the fraction is close to $sqrt2$. Similarly, when $a_n$ is very large, the fraction is very close to $a_n/sqrta_n^2=1$. In between these extremes, it is continuous, hence bounded. You could build that into a formal proof, but that is clearly overkill! But this quick reasoning gives you the answer very easily, it is for sure bounded.

Instead, note that you need to prove the existence of some $M$ so that $$ 2+a_n le Msqrt2+a_n^2 .$$
Now square that inequality (note that both sides are positive), rearrange it a bit, say by collecting equal powers of $a_n$, and try find some value of $M$ that lets you prove it.

Alt. hint:   let $a = sqrt2 tan t$, then:

$$
frac2+asqrt2+a^2 = fracsqrt2sqrt2cdot fracsqrt2 + tan tsqrt1+tan^2 t = (sqrt2+tan t) cdot |cos t| ;le; sqrt2|cos t| + |sin t|
$$