proof by definition of limit of $\lim_{x\to -\infty}\tanh(x)=-1$

I am trying to show this limit $$\lim_{x\to -\infty}\tanh(x)=-1$$ by definition I must have to $\forall \epsilon>0 \exists c \forall x<c \rightarrow |f(x)-L|<\epsilon$.

My attempt $$|\tanh(x)+1|<\epsilon$$ $$\tanh(x)<\epsilon-1$$

I don't know if I can take $c=tanh^{-1}(\epsilon-1)$ or $c=min\{-1,tanh^{-1}(\epsilon-1)\}$, I feel quite confused, any help I will appreciate

Solution 1:

Let $f(x)=\tanh(x)$. To show that $\lim_{x\to-\infty}\tanh(x)=-1$, we will show that for every $\epsilon>0$, there exists $c\in\mathbb{R}$ such that $|f(x)-(-1)|<\epsilon$ whenever $x<c$.

Let $\epsilon>0$. Choose a negative number $c<0$ large in absolute value so that $x<c$ implies $$e^{2x}=\frac{1}{e^{-2x}}<\frac{1}{e^{-2c}}<\epsilon.$$ Then, whenever $x<c$, we have $$|f(x)-(-1)|=\Big|\frac{e^{2x}-1}{e^{2x}+1}+1\Big|=\frac{e^{2x}}{e^{2x}+1}<\frac{\epsilon}{1}=\epsilon,$$ finishing the proof.