So i got one quest if im correct. U Will have 2 case because of absolute value. F(x)>0 and f(x)<0. What i want to say is when its case 2 i Will solve ur way and case 1 is what i did just do. Then im pretty much confused what to do with these two case
The problem says to find N from the definition of the limit. It must be true that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ for $\displaystyle x > N$ and $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>1.5$ for $\displaystyle x > N$. The graph shows that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ for all $\displaystyle x > 0$. Solving the second inequality, you find $\displaystyle N > 0$ such that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>1.5$ for $\displaystyle x > N$. So, if $\displaystyle x > \max(0,N)$, we have both $\displaystyle x > 0$ and $\displaystyle x > N$ and therefore both inequalities are true. But $\displaystyle \max(0,N) = N$, so both of those inequalities are true for $\displaystyle x > N$.
Hello Emakarov!
If this would be on exam would this be a good answer(i will skip the math part):
By the graph or calculate we can se to right side of x intercept (0) we can se that it will be <0 so we only will get positive x intercept if we use x<0. So i calculate the x intercept when -(f(x)-L)<epsilon (0.5) and get x intercept as 2.82 and 0.2528. then i can set like 3 in function and look if its lower then epsilon (0.5) and it is.
Sorry, this is very hard to read.
"we can se that it will be <0": what will be < 0?
"we only will get positive x intercept": why are you talking about x-intercepts? And x-intercepts of what? The x-intercepts of the original function
$\displaystyle f(x)=\frac{\sqrt{4x^2+1}}{x+1}$ (*)
do not arise in this problem at all.
"the x intercept when -(f(x)-L)<epsilon": you can't talk about the x-intercept "when" an equation holds. The concept of an x-intercept is only applicable to a function, not to an inequality. An inequality has solutions, which is a possibly infinite set of real numbers. Sometimes this set can be expressed using several inequalities of the form x > ... or x < ... .
"and get x intercept as 2.82 and 0.2528": the second value should be negative, but it is not important here.
"then i can set like 3 in function and look if its lower then epsilon (0.5) and it is": "like" is not appropriate in mathematical text. Which function: f(x) from (*) above or |f(x) - 2|? How can you check whether |f(x) - 2| < 0.5 for all x > 3, i.e., for infinitely many x? And why would you need to check this if you have just solved this inequality?
ima try do my best to explain now, im not a good explainer :/
when i mean <0 then i mean with solving absolute value equation we will have positive value when -(f(x)-L)
tbh idk how to describe this with words. I basicly understand not really good on this but i need to read more about this.
You may have noticed how few of us are now replying to your post.
I will not as long as you are so discourteous as to posting unreadable images.
If I were you, I would get access to a PC or a tablet computer.
I have a 10in tablet that is a fully function cell phone.
It is easy to use on this site.
You could use Tapatalk on your phone.
OR you could learn to use the camera correctly.
It appears that you are just too lazy to do that.