a stupid question about function's limit Ok so recently I was doing a problem about limit. I know that every rational function that has a same exponent for the numerator and denominator will result in a definite number( I dont know if my wording is right but I just mean $ \lim_{x\rightarrow \infty }\frac{\sqrt{x^{2}+2}}{x}=1$. However I was thinking about the range: if this particular function wont actually head become exactly $1$ and it is describing the ratio of a sequence, will a number in the sequence become infinite large because the ratio is always larger than one? Or is there a particular number that the number in the sequence can't exceed( I actually will be gald to see if there is a maximum ratio and the number in the sequence cannot exceed a particular number...)? If someone can answer me I will be really glad...(a stupid precal student wrote this so plz dont judge me)
Sorry i have extremely bad wording but I mean if this function is only describing the ratio between elements in a set and the elements itself is dependent on the function( new element is the ratio times previous element) and the function is dependent on the elements( the x value in the ratio function is a element in the set that has the "latest value")will the set have infinitely many numbers and their quantitative value will head toward infinite since the ratio is always a little greater than one?
 A: Well, at $x = 0$ the function you describe goes to infinity - so for very small $x$, $\frac{\sqrt{x^2+2}}{x}$ can be as large as you like.
But since the limit does go to $1$, this behavior eventually stops. To say this precisely: for any $\epsilon > 0$, there is some point after which $\frac{\sqrt{x^2 + 2}}{x}$ never again exceeds $1 + \epsilon$. For example, when $x > 1$, $\frac{\sqrt{x^2 + 2}}{x}$ never exceeds $\sqrt{3}$.
EDIT: Based on your comments, it sounds like you're asking about a sequence where if one term is $x$, the next is $\sqrt{x^2 + 2}$. Beginning from $1$, this sequence goes $1, \sqrt{3}, \sqrt{5}, \sqrt{7}, \ldots$. As you can see, what's going on is that (1) squaring removes the square root, then (2) we add $2$, then (3) we put the square root back. But that means the number under the radical is increasing by $2$ each time - so it gets infinitely large, and therefore this sequence gets infinitely large.
As a general rule, this is a good approach to sequences; calculate the first five or six terms and see if you spot a pattern, then try to explain why that pattern was there.
Importantly, although this sequence tends toward $\infty$, it's not because the ratio is always a little more than $1$. In the sequence $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots$, the ratio is always slightly more than $1$, but the sequence itself converges to $1$ instead of going to $\infty$.
