Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$

Question

Formal proof for the limit of $\frac{\tanh(x)-1}{e^{-2x}}$ as $x \rightarrow \infty$.

So far

Keep in mind I have to use the definition for a limit. I.e for this would be a proof for the limit at $\infty$ with a finite value (I have checked this to be $-2$). I would have to use the definition that states $\lim_{x \ \to \ \infty} f(x) = L \Leftrightarrow \forall \ \epsilon>0\; (\exists \ \delta : (\;x>\delta\implies |f(x) - L|\leq\epsilon)). $ i.e I start off by by finding that $$\left | \frac{\tanh{x}-1}{e^{-2x}}-(-2) \right | \iff \left | \frac{2}{1+e^{2x}} \right |$$ Or just $\frac{2}{1+e^{2x}}$ since we will assume $\epsilon >0$. Now I would have to to start my proof and this is where I struggle.

I know that "for alle $\epsilon>0$ let $\delta=\frac{2}{1+e^{2\epsilon}}>0$ and $x>\delta>0$....."

*Remark: I know we use $N,M,K$ and not $\delta$ normally.

Answer 1

Based on feedback in comments I will add reasonable detail in the answer. Some of it may be skipped by those already well versed.

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The definition of $\lim_{x\to\infty} f(x) =L$ requires us to show that for every $\epsilon >0$ we can find a corresponding $\delta>0$ such that the inequality $|f(x) - L|<\epsilon $ holds whenever $x>\delta$.

But this does not mean that we start with inequality $|f(x) - L|<\epsilon$ and manipulate it according to valid laws to transform it into "$x>\text{something} $" and then put $\delta=\text{something} $. This approach is what I call solving an inequality and I request readers never to do this while giving formal proofs of limits.

Instead what really needs to be done is to find an expression, say $g(x) $, which is very simple in form and satisfies $|f(x) - L|<g(x) $ and then try to solve $g(x) <\epsilon$. The simple form of $g(x) $ is needed to make this solving part trivial.

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Here the desired limit $L=-2$ and we need to use the fact that $e^x>1+x$ if $x>0$ in order to find a suitable $g(x) $. By using it you get $$|f(x) - L|=\frac{2}{1+e^{2x}}<\frac{2}{2x+2}=\frac{1}{x+1}<\frac{1}{x}$$ Now the rightmost expression is less that $\epsilon$ if $x>1/\epsilon$. Thus we can take $\delta$ to be any number greater than or equal to $1/\epsilon$ (in particular $\delta=1/\epsilon$ works).

Essentially you have to do a very gross estimation of $|f(x) - L|$ by $g(x) $. The goal to have a very very simple form for $g(x)$ and you don't need to worry however gross the estimation by $g(x)$ is.

Answer 2

$$\frac{\tanh(x)-1}{e^{-2x}}=\bigg(\frac{e^{2x}-1}{e^{2x}+1}-1\bigg)\cdot e^{2x}=\frac{-2e^{2x}}{{e^{2x}+1}}$$

So $$\bigg|\frac{-2e^{2x}}{e^{2x}+1}-(-2)\bigg|=\bigg|2-\frac{2e^{2x}}{e^{2x}+1}\bigg|=\frac{2}{e^{2x}+1}$$

Now if $\varepsilon>0$ find $\delta>0$ such that when $x>\delta$ we have that $e^{2x}+1>\frac{2}{\varepsilon}$. We can do that because the function $\phi(x)=e^{2x}+1$ tends to $\infty$ as $x\to\infty$. Then for $x>\delta$ we have that $$e^{2x}+1>\frac{2}{\varepsilon}\iff\frac{2}{e^{2x}+1}<\varepsilon. $$