I faced recently this exercise:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f$ is strictly increasing and the following hold: $f^{-1}(x) \leq x-1 , x \in \mathbb{R}$ and $\lim_{x\rightarrow 0} f^2(x)=1$. Prove that $(a)$ $f(x)\geq x+1$ holds for all $x \in \mathbb{R}$ and $(b)$ $\lim_{x\rightarrow 0} f(x)=1$
By solving the exercise I realized that $f$ is not necessarily strictly increasing. I mean that if we were given "$f$ is $1-1$" instead, it would suffice to solve the exercise.
The solution that the author suggests for $(a)$ is the following:
The inequality $f^{-1}(x) \leq x-1$ holds for all $x \in \mathbb{R}$ therefore since $f$ is strictly increasing we get $f(f^{-1}(x))\leq f(x-1) \Rightarrow f(x-1)\geq x$. Let $x \rightarrow x+1$ thus $f(x)\geq x+1 , x \in \mathbb{R}$
But as I said, I claim that $(a)$ can be solved without using the fact that $f$ is strictly increasing such that:
The inequality $f^{-1}(x)\leq x-1$ holds for all real $x \in \mathbb{R}$. So let $x \rightarrow f(x)$ getting $f^{-1}(f(x))\leq f(x)-1 \Rightarrow f(x)\geq x+1$. But $f(x) \in \mathbb{R} \forall x \in \mathbb{R}$ thus our new inequality holds for all real $x$.
Is my solution wrong?
Now as far as $(b)$ is concerned, I did the following:
It is $\lim_{x\rightarrow0}f^2(x)=1 \Leftrightarrow \lim_{x\rightarrow 0} |f(x)|=1$
But $f(x)\geq x+1$ therefore $f(x)>0 \forall x >-1$ thus $f(x)>0 $ for all $x$ "near" zero, which implies $|f(x)|=f(x)$ for all $x$ "near" zero. Thus $\lim_{x\rightarrow0} f(x)=1$.
I'm not sure the above is rigorous enough though.
The solution that the author suggests is: $x+1 \leq f(x) \leq \dfrac{f^2(x)+1}{2}$ and by using the squeeze theorem concludes that $\lim_{x\rightarrow0}f(x)=1$
Please include answers that do not involve the ε-δ definition of a limit.