how to prove that $f(x) = \ln (x^2 +1 )$ is not uniform continuous?

Question

$f:[0, +\infty) \to \Bbb{R}$

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It is the last function in the photo. I couldn't find two Sequences that will prove that and also couldn't find a proof for saying that because $x^2 +1$ is not uniformly continuous, therefore $\ln( x^2 +1)$ is also not uniformly continuous

Answer 1

Let $ f(x)=log(1+x^2)$. Note that $$|f'(x)|=|2x/(1+x^2)|\le 1$$ for all real numbers $x$. Therefore $|f(x)-f(y)|=|f'(c)(x-y)|\le |x-y|$. Thus f(x) is uniformly continuous.

Answer 2

But it is! $$ |\log(1+x^2)-\log(1+y^2)|\stackrel{\text{mean value theorem}}\leq |x-y|\sup_{t\in [0,\infty)}\left|\frac{2t}{1+t^2}\right|=|x-y| $$