Question about limits at infinity: $f(x) < g(x)$ and $f(x) < \epsilon$ imply $g(x) < \epsilon$? I am going over the proof of $\lim\limits_{x \rightarrow \infty} \frac{x^2+2x}{2x^2+1} = \frac{1}{2}$. Let $f(x) = \frac{x^2+2x}{2x^2+1}$.
After setting $\epsilon >0$, we have $|f(x) - \frac{1}{2}| = |\frac{x^2+2x}{2x^2+1} - \frac{1}{2}| = |\frac{4x-1}{2(2x^2+1)}| < \epsilon$. We find that if $x > \frac{1}{4}$, then $\frac{4x-1}{2(2x^2+1)} < \epsilon$.
Now, the argument goes something like:
$$f(x) = \frac{4x-1}{2(2x^2+1)} < \frac{2x}{2x^2+1} < \frac{2x}{2x^2} = \frac{1}{x} = g(x)$$
And then it concludes that $f(x) < \frac{1}{x} = g(x) < \epsilon $ so $x > \frac{1}{\epsilon}$, and finally take $\max(\frac{1}{4}, \frac{1}{\epsilon})$ to prove the claim.
We have shown $f(x) < g(x)$ and we had $f(x) < \epsilon$. Why is it necessarily true that $g(x) < \epsilon$? Because in general if $a < b$ and $a < c$, we cannot conclude that $b < c$, since it is also possible that $c \leq b$. The inequality $g(x) < \epsilon$ does not look fully justified to me as is, especially because we have inequalities in the intermediate steps between $f(x)$ and $g(x)$ and not equalities.
 A: The proof you are quoting seems to redefine $f(x)$. Worse, it is missing some connecting words to flesh out the argument. Some of the '<' you are seeing are aspirational: we want to assert inequality, but the inequality isn't true without additional conditions on $x$. The rest of the proof seeks to establish those conditions.
Here's how the proof should be read, with the connecting words filled in:  First, we are given $\epsilon>0$. We seek a threshold $M$ so that if $x>M$, then $|f(x)-\frac12|<\epsilon$. Using algebra, we get
$$\left|f(x)-\frac12\right| = \left|\frac{4x-1}{2(2x^2+1)}\right|.$$ We want to arrive at the assertion $\left|\frac{4x-1}{2(2x^2+1)}\right|<\epsilon$. It isn't true for all $x$, so we seek conditions to make this assertion true. But the absolute value is getting in the way. We notice that if $x>\frac14$, then the numerator $4x-1$ is positive so we can drop the absolute value:
$$
\text{If $x>\frac14$, then}\qquad\left|f(x)-\frac12\right|=\frac{4x-1}{2(2x^2+1)}.\tag1$$
Next, we try to simplify the RHS of (1) to more easily see how to get it below $\epsilon$. Using algebra, we see
$$\frac{4x-1}{2(2x^2+1)}<\frac1x\tag2$$ for all $x$. So provided $\frac1x<\epsilon$, then the RHS of (2) will be less than $\epsilon$ and hence the LHS of (2) will be less than $\epsilon$ and hence the LHS of (1) will be less than $\epsilon$ (provided we also have $x>\frac14$). In summary: If $x>\frac14$ and $x>\frac1\epsilon$, then $|f(x)-\frac12|<\epsilon$. So we can take the threshold $M:=\max(\frac14, \frac1\epsilon$), proving the claim.
A key aspect of the proof is that we are seeking an upper bound on (1), so in (2) we must establish a chain of '<' inequalities, leading to the simple upper bound $\frac1x$. A chain of '>' inequalities, or a mixture of '>' and '<', won't get us anywhere. We can summarize the algebra in one line, with caveats:
$$\left|f(x)-\frac12\right|=\left|\frac{4x-1}{2(2x^2+1)}\right|\stackrel{(1)}=\frac{4x-1}{2(2x^2+1)}\stackrel{(2)}<\frac1x\stackrel{(3)}<\epsilon
$$ where the caveats are: (1) provided $x>\frac14$
(3) provided $x>\frac1\epsilon$. Note assertion (2) is true for all $x$.
