# Thread: epsilon delta surroundings question

1. ## epsilon delta surroundings question

i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}

2. Please post the entire question.

3. prove by limit definition that
x/(x+1)=1
x->infinity

4. To show any limit of the form $\displaystyle \displaystyle \lim_{x \to \infty}f(x) = L$

then you would have to show

$\displaystyle \displaystyle x > N \implies |f(x) - L| < \epsilon$.

Therefore, to show that $\displaystyle \displaystyle \lim_{x \to \infty}\frac{x}{x+1} = 1$

you will need to show that

$\displaystyle \displaystyle x > N \implies \left|\frac{x}{x + 1} - 1\right| < \epsilon$ where $\displaystyle \displaystyle N$ is a positive integer.

So solve $\displaystyle \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon$ for $\displaystyle \displaystyle x$

$\displaystyle \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon$

$\displaystyle \displaystyle \left|\frac{x}{x + 1} - \left(\frac{x + 1}{x + 1}\right)\right| < \epsilon$

$\displaystyle \displaystyle \left|-\frac{1}{x + 1}\right| < \epsilon$

$\displaystyle \displaystyle \frac{1}{|x + 1|} < \epsilon$

$\displaystyle \displaystyle |x + 1| > \frac{1}{\epsilon}$

$\displaystyle \displaystyle x + 1 < -\frac{1}{\epsilon}$ or $\displaystyle \displaystyle x + 1 > \frac{1}{\epsilon}$

$\displaystyle \displaystyle x <- \frac{1}{\epsilon} - 1$ or $\displaystyle \displaystyle x > \frac{1}{\epsilon} - 1$.

So choose $\displaystyle \displaystyle N = \frac{1}{\epsilon} - 1$.

Proof: If $\displaystyle \displaystyle N = \frac{1}{\epsilon} - 1$ and $\displaystyle \displaystyle x > N$, then

$\displaystyle \displaystyle x > \frac{1}{\epsilon} - 1$

$\displaystyle \displaystyle x + 1 > \frac{1}{\epsilon}$

$\displaystyle \displaystyle |x + 1| > \frac{1}{\epsilon}$

$\displaystyle \displaystyle \frac{1}{|x + 1|} < \epsilon$

$\displaystyle \displaystyle \left|\frac{1}{x + 1}\right| < \epsilon$

$\displaystyle \displaystyle \left|\frac{x}{x + 1} - 1\right| < \epsilon$

$\displaystyle \displaystyle \left|f(x) - L\right| < \epsilon$.

Since $\displaystyle \displaystyle x > N \implies |f(x) - L| < \epsilon$, that means $\displaystyle \displaystyle \lim_{x \to \infty}\frac{x}{x + 1} = 1$.

5. ok i tried to prove the same thing with
$\displaystyle lim_{x\rightarrow\infty}\sqrt{2x-sin3x}=\infty$
my book says that this limit would work only if
for every N>0 we could find M>0 so for every x>M f(x)>N would work

so i need to find M for which $\displaystyle \sqrt{2x-sin3x}$>N would work
x>M

this is as far as i could go
there is no absolute value innequality for me to play with like you did previosly

$\displaystyle \displaystyle \sqrt{2x - \sin{3x}} > N$

$\displaystyle \displaystyle 2x - \sin{3x} > N^2$

$\displaystyle \displaystyle |2x - \sin{3x}| > |N^2|$

$\displaystyle \displaystyle |2x - \sin{3x}| > N^2$

Now, from the triangle inequality

$\displaystyle \displaystyle |2x - \sin{3x}| = |2x + (-\sin{3x})|$

$\displaystyle \displaystyle |2x + (-\sin{3x})| \leq |2x| + |-\sin{3x}|$

$\displaystyle \displaystyle |2x + (-\sin{3x})| \leq 2|x| + |\sin{3x}|$

$\displaystyle \displaystyle |2x + (-\sin{3x})| \leq 2|x| + 1$ since $\displaystyle \displaystyle |\sin{3x}| \leq 1$.

Since $\displaystyle \displaystyle |2x - 3\sin{x}| >N^2$ and $\displaystyle \displaystyle 2|x| + 1 \geq |2x - \sin{3x}|$ that means

$\displaystyle \displaystyle 2|x| + 1 > N^2$

$\displaystyle \displaystyle 2|x| > N^2 - 1$

$\displaystyle \displaystyle |x| > \frac{N^2 - 1}{2}$.

So we can choose $\displaystyle \displaystyle M = \frac{N^2 - 1}{2}$.

7. ok but like you did previosly this is only a sketch i do for my self to find the M.

the actual proof starts with : (i tried to do it like you did)
there is N for which we pick M to be $\displaystyle M=\frac{N^2-1}{2}$
so $\displaystyle x>\frac{N^2-1}{2}$ thus saying that x>0
$\displaystyle 2x>N^2-1$
$\displaystyle 2x+1>N^2$
$\displaystyle \sqrt{2x+1}>N$

what now , here i proved the limit for $\displaystyle \sqrt{2x+1}$ not for the original limit.
how i connect it to the original limit
?