Hello i got problem with nr 62. I got some progress but i dont understand
Yes, for ε = 0.5, N = 5 and even N = 3 works.
So, you need to determine for which x > 0 it is the case that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$. Multiply both sides by (two times) the denominator, take square of both sides and solve (approximately) the resulting quadratic equation.
You start with the quadratic inequality, which skips a lot of steps. I don't agree with the inequality, but I can't point out the step with an error because these steps are skipped. I wrote a couple of intermediate results in post #6. I also recommended multiplying both sides by 2 (or 4) to get rid of fractions like 1.5.
Here is a sequence of steps I recommend. Start with $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>1.5$.
1. Multiply both sides by (x + 1). (The direction of the inequality does not change because we are looking for x > -1, where x + 1 > 0.)
2. Multiply both sides by 2 to get rid of 1.5.
3. Take square of both sides. (This could lead to appearance of spurious solutions, but we ignore this for now.)
4. Move everything to the left-hand side and add like terms.
5. Solve the quadratic equation obtained when > is replaced with =. Let's denote the solutions by $\displaystyle x_1$ and $\displaystyle x_2$ where $\displaystyle x_1 < 0$ and $\displaystyle x_2 > 0$.
6. Since the leading coefficient of the quadratic polynomial f(x) is positive and the inequality has the form f(x) > 0, the solutions to the inequality are $\displaystyle x < x_1$ and $\displaystyle x > x_2$. We are interested in $\displaystyle x_2$.
Edit: Sorry, I missed that there are two attached images and not one. I'm looking at the second one...
OK, I see that you started from $\displaystyle \left|\frac{\sqrt{4x^2+1}}{x+1}-2\right|<0.5$, which you converted into $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}-2<0.5$. In fact, if we denote $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}$ with z, $\displaystyle |z-2|<0.5$ is equivalent to $\displaystyle -0.5<z-2<0.5$. Adding 2 to all sides, we get $\displaystyle 1.5<z<2.5$. So you are right that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ is a part of the original inequality. However, if you saw the graph from post #2, you must have seen that as x tends to infinity, the graph approaches 2 from below. This means that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2$ for positive x, so the inequality $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ is automatically true for positive x. What we are interested in is the other part: $\displaystyle 1.5<\frac{\sqrt{4x^2+1}}{x+1}$. It becomes true only starting from some N > 0. I assumed you saw all this after post #2 and therefore I recommended solving
$\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$ (*)
in post #4. Later, in post #6, I wrote a couple of inequalities you obtain while solving (*). You ignored all this and started solving a different inequality.