Prove $\lim_{x \to 0^{-}}\frac1{1 + e^{1/x}}=1$ by $\epsilon - \delta$ - definition

Question

how to prove by $\epsilon - \delta$ - definition that $$\lim_{x \to 0^{-}}\frac1{1 + e^{1/x}}=1.$$

My attempt: I arrived at $\dfrac1{e^{1/x} + e^{2/x}}< \epsilon$ and I know that $-\dfrac 1\delta > \dfrac 1x > 0$ by our assumption, hence $e^{1/\delta} < 1/e^{1/x} < 1$ and also $e^{2/\delta} < 1/e^{2/x} < 1$ , finally I arrived that $\dfrac1{e^{1/x} + e^{2/x}}< 1/2$ am I correct? then I will take $\delta \leq 1/2$ correct?

Answer 1

Let $0<\epsilon <1$ and $\delta =\frac 1 {log \, (\frac 1 {\epsilon})}$. Then $x<0$ and $0<|x| <\delta$ implies $-\delta <x <0$ so $e^{\frac 1 x} <\frac { \epsilon} {1-\epsilon}$ so $\frac {e^{\frac 1 x}} {1+e^{\frac 1 x}} <\epsilon$. Observe that $|\frac 1 {1+e^{\frac 1 x}}-1|=\frac {e^{\frac 1 x}} {1+e^{\frac 1 x}} <\epsilon$. How does one arrive at this $\delta $?: We want $\frac {e^{\frac 1 x}} {1+e^{\frac 1 x}} <\epsilon$ and the inequality $\frac x {1+x} < \epsilon $ is same as $x < \frac {\epsilon} {1-\epsilon}$ as seen by a simple algebraic manipulation. Now you can see how I got my $\delta$.

Answer 2

Then what shall I do to find $\delta$?

I'm not sure if I understand you correctly but as far as I understand you, you want to find the largest possible value of $\delta$.

This is however not required!

If you find out that $\delta = \frac{1}{2}\epsilon$ is a valid solution for $(\lim\limits_{x \to ...} 2x)$ then $\delta = \frac{1}{1000}\epsilon$ is also a valid solution:

If some condition is true for all $x \in ]0;\frac{1}{2}\epsilon[$ then the condition is of course also true for all $x \in ]0;\frac{1}{1000}\epsilon[$.

I arrived at $\dfrac1{e^{1/x} + e^{2/x}}< \epsilon$

Now you can continue with:

$\dfrac1{e^{1/x} + e^{2/x}} < \dfrac1{e^{1/x}} < \epsilon$

If you find some $\delta$ that satisfies $\dfrac1{e^{1/x}}<\epsilon$ for all $x \in ]0;\delta[$ this $\delta$ will also satisfy $\dfrac1{e^{1/x} + e^{2/x}} < \epsilon$ for all $x \in ]0;\delta[$.

$\delta = \dfrac 1 {-\ln\epsilon}$ should be a valid solution (assuming $\epsilon < 1$).

Edit

How did you arrived at this choice?

The idea here is the following one: You are searching for some value $\delta$ which satisfies $f(x)<\epsilon$ for all $x \in ]0;\delta[$.

Unfortunately the function $f(x)$ (in your case: $f(x)=\dfrac1{e^{1/x} + e^{2/x}}$) is rather complicated so it is very difficult to find a suitable $\delta$.

The idea is now to find a much simpler function $g(x) > f(x)$ and to find a $\delta$ which satisfies $g(x)\leq\epsilon$ for all $x \in ]0;\delta[$.

Then $\epsilon \geq g(x) > f(x)$ for all $x \in ]0;\delta[$ so the value $\delta$ you found also satisfies your first condition ($f(x)<\epsilon$).

Why I have chosen $g(x)=\dfrac{1}{e^{1/x}}$?

It was looking at $f(x)$ and searching a function that satisfies $g(x)>f(x)$ which was rising monotonously and where I was able to calculate $g^{-1}(x)$.

Why this choice of $\delta$ valid?

I'm assuming that the following properties of the function $e^x$ are known (and don't have to be proven):

• $e^x>0$ for all values of $x\in\mathbb R$
• $e^x$ is rising monotonously
• There is a function called "$\ln$" here that satisfies $e^{-\ln\epsilon}=\frac{1}{\epsilon}$ for all $\epsilon \in ]0;1[$

$e^{2/x}>0 \Rightarrow \\ e^{1/x} + e^{2/x} > e^{1/x} \Rightarrow \\ f(x) = \dfrac1{e^{1/x} + e^{2/x}} < \dfrac1{e^{1/x}} = g(x)$

$e^x$ is rising monotonously. This means (assuming $x>0$):

$x<\delta \Rightarrow \\ 1/x > 1/\delta \Rightarrow \\ e^{1/x} \geq e^{1/\delta} \Rightarrow \\ g(x) = \dfrac{1}{e^{1/x}} \leq \dfrac{1}{e^{1/\delta}} = g(\delta) = g(\dfrac 1 {-\ln\epsilon}) = \epsilon$

So $f(x) < g(x) \leq g(\delta) = \epsilon$ for all $0 < x < \delta$.