Epsilon delta proof with x going towards infinity.

Question

I have the following function: $h(x)=\frac{-2e^x}{e^x+\frac{1}{e^x}}$ with $x \rightarrow \infty$ and $h(x) \rightarrow -2$ and I am attempting to prove it with an epsilon delta proof.

I then do the following:

$\Big\lvert h(x)-(-2)\Big\rvert=\dfrac{-2e^x+2(e^x+\dfrac{1}{e^x})}{e^x+\dfrac{1}{e^x}}=\dfrac{\dfrac{2}{e^x}}{e^x+\dfrac{1}{e^x}}<\epsilon$

and attempt to create an upper bound with the following inequality and evolving around it:

$e^x>1+x \Rightarrow e^x+\frac{1}{e^x} > 1+x+\frac{1}{e^x} \Rightarrow \frac{1}{e^x+\dfrac{1}{e^x}}<\dfrac{1}{1+x+\dfrac{1}{e^x}} \Rightarrow \frac{\dfrac{2}{e^x}}{e^x+\dfrac{1}{e^x}}<\dfrac{\dfrac{2}{e^x}}{1+x+\dfrac{1}{e^x}}$

But can't seem to get further. Help would be appreciated!

Answer 1

As you said $\dfrac{\dfrac{2}{e^x}}{e^x+\dfrac{1}{e^x}}<\epsilon$ That's equivalent to saying $\frac{2}{e^{2x}+1}<\epsilon$ so $e^{2x}$+1>$\frac{2}{\epsilon}$, so $e^{2x}>\frac{2}{\epsilon}-1$. Now, if you have a negative/ zero on the right hand side, you can choose h=0. Else, you can pick h = $ 1/2 *ln(\frac{2}{\epsilon}-1)$ so x>h $\implies |h(x)-(-2)|<\epsilon$

so satisfies that limit.

As John Douma has stated above, this is not technically epsilon delta : you look for some h s.t x>h implies $|h(x)-(-2)|<\epsilon$ for this type of limit.

Answer 2

You need to be cautious: What you want to show is that $$|h(x) - (-2)| < ε$$ for $x$ large; this is not what you are given!

To do this, fix $ε>0$. We have $$|h(x)-(-2)| = \frac{\frac{2}{e^x}}{e^x + \frac{1}{e^x}} = \frac{2}{e^{2x} + 1} =: f(x).$$ If $ε\geq 1$, choose $M = 0$. we immediately have that $f(x)\leq 1 \leq ε$ for any $x\geq M = 0$, since $e^{2x} + 1 \geq 2$.
If $0<ε<1$, choose $M = \frac{1}{2}\log(\frac{2}{ε} - 1)$. Then, by monotonicity of $f$ (check this!), we have for all $x\geq M$ that $$f(x) \leq f(M) = \frac{2}{\frac{2}{ε} - 1 + 1} = ε$$ as desired.

Answer 3

You need to show that for $\epsilon >0 $ there is an $M$, such that $x > M$ implies $|h(x) - (-2)| <\epsilon.$ (Comment by J. Douma)

Note:

Let $x>0:$

$|h(x) - (-2)|=\dfrac{2}{e^{2x}+1}<\dfrac{2}{1+2x+1}<\dfrac{1}{x}.$

Choose $M>1/\epsilon$. Then

$x>M$ implies

$|h(x) - (-2)|< 1/x<1/M <\epsilon.$

Used: $e^{2x}=1+(2x)/1! + (2x)^2/2!...$