- #1

mathmari

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MHB

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We consider the function $f(x)=\tanh x$.

I shown so far the following:

- The function is defined for all $x\in \mathbb{R}$, i.e. $D_f=\mathbb{R}$.
- The function is strictly increasing.
- $\displaystyle{\lim_{x\rightarrow -\infty}\tanh x=-1}$ and $\displaystyle{\lim_{x\rightarrow \infty}\tanh x=1}$. And since $\tanh$ is continuous on the whole $\mathbb{R}$ the range is $(-1,1)$.

Now I want to show that $\tanh$ is uniformly continuous on $\mathbb{R}$ (without using differential calculus).

Could you give me a hint how we could show that? Do we have to use the definition? (Wondering)