# Math Help - Epsilon nr 2

1. ## Epsilon nr 2

Hello this is almost same problem like last one. (ima try my best with latex)
for the limit
lim x-> -infinty $\frac{\sqrt{4x^2+1}}{x+1}=-2$*
illustrade Defination 8 by finding values of N that correspond to $\varepsilon=0.5$ and $\varepsilon=0.1$

So basicly what i have done:
1. i start to set $\frac{\sqrt{4x^2+1}}{x+1}+2=0$ 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 $\frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l if $x (what i mean with l $\frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l that "l" is absolute value sorry idk how to use it with latex.)
so i get the equation $-9x^2-50x-21>0$ 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

2. ## Re: Epsilon nr 2

Originally Posted by Petrus
(ima try my best with latex)
Bravo!

Originally Posted by Petrus
So basicly what i have done:
1. i start to set $\frac{\sqrt{4x^2+1}}{x+1}+2=0$ and get x=-0.125
This equation has no solutions. The graph shows that $f(x)=\frac{\sqrt{4x^2+1}}{x+1}$ does not take values between -2 and 0, inclusive, for any value of x. It is also not clear why you need to solve this equation.

Originally Posted by Petrus
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.
This is not clear at all. Which lower number? Why do you consider x = -2?

Originally Posted by Petrus
2. i set that l $\frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l if $x (what i mean with l $\frac{\sqrt{4x^2+1}}{x+1}+2<0.5$l that "l" is absolute value sorry idk how to use it with latex.)
The absolute value should end after "+2" and before "<", not after "0.5". Vertical bar "|" is used to write absolute value in LaTeX. To make the bars the same vertical size as the formula between them, use \left| ... \right| instead of just | ... |.

Originally Posted by Petrus
so i get the equation $-9x^2-50x-21>0$ 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
Solving the inequality here is a little tricky because one has to multiply by a negative number (which changes the direction of the inequality) and take square of both sides (which introduces extra roots). Therefore, it is better to look at the graph and determine the equation to solve. The red graph of f(x) must be inside the blue band of width 1 (i.e., 2 * 0.5) around the line y = -2. Intersection of which lines do you need to determine?

In fact, the equation $-9x^2-50x-21=0$ is correct, and the relevant root is approximately -5.10. Therefore, if you are looking for an integer N, it would be -6.

3. ## Re: Epsilon nr 2

Hello Emakorv.
Im starting to geting really confused because i asked a math teacher and he toled me " if u smart.." then u would look for the roots of the equation in this one the x go -infinity and then i can take a number less then that root( i got the root -0,125 and if i take -1 it will be -1+1 on bottom with other words i get zero on bottom of division.... so i choose -2.. with other words if its f(x)>0 i would use the possitive only and if its f(x)<0.. this one is killing me i have been reading about this for 2 days.. What i understand just says im wrong... I dont know what to do...

edit: its nothing about it havs to be in range of epsilon just look for the graph and look for roots.. (what i said on step 1. i hope u understand what i mean and i hope if u can confirme if its true)
I would like to solve this problem without looking at the graph, i dont wanna get stick to the graph.

4. ## Re: Epsilon nr 2

Originally Posted by Petrus
Im starting to geting really confused because i asked a math teacher and he toled me " if u smart.." then u would look for the roots of the equation in this one the x go -infinity and then i can take a number less then that root( i got the root -0,125
First, the second root of $9x^2 + 50x + 21 = 0$ is -0.46. Second, it is the wrong root that appears because of taking square of both sides. You may check that $\frac{\sqrt{4x^2+1}}{x+1}$ is significantly less than -2.5 at x = -0.46, i.e., $\left|\frac{\sqrt{4x^2+1}}{x+1}-(-2)\right|$ is nowhere close to 0.5.

Originally Posted by Petrus
edit: its nothing about it havs to be in range of epsilon just look for the graph and look for roots.. (what i said on step 1. i hope u understand what i mean and i hope if u can confirme if its true)
I would like to solve this problem without looking at the graph, i dont wanna get stick to the graph.
I don't think you currently have the skills to solve the inequality $\left|\frac{\sqrt{4x^2+1}}{x+1}-(-2)\right|<0.5$ purely algebraically. I recommend looking at the graph, determining the equation you need to solve and then solving it.

5. ## Re: Epsilon nr 2

U are right! One quest i do same again for next one with epsilon 0.1 and i should get n=-22?
And u are true thats why i could use graph to this problem but i always try not to use graph ohh well, i Will learn that later