Hello this is almost same problem like last one. (ima try my best with latex)

for the limit

lim x-> -infinty$\displaystyle \frac{\sqrt{4x^2+1}}{x+1}=-2$*

illustrade Defination 8 by finding values of N that correspond to $\displaystyle \varepsilon=0.5$ and $\displaystyle \varepsilon=0.1$

So basicly what i have done:

1. i start to set $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}+2=0$ and get x=-0.125 and with this i can basicly take a lower number put it on orginal function (*). Basicly when u calculate with absolute value u get 2 case and one is when f(x)>0 and f(x)<0 so when i put x=-2 on (*) then i get a negative result that means i only need to calculate with the case f(x)<0.

2. i set that l$\displaystyle \frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l if $\displaystyle x<N$ (what i mean with l$\displaystyle \frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l that "l" is absolute value sorry idk how to use it with latex.)

so i get the equation $\displaystyle -9x^2-50x-21>0$ and when i solve this equation i get that x_1=5 and x=0.457

What shall i do next?Facit says to my book the answer shall be N≤-6, N≤-22