Existence of $x \in [0, 2]$ such that $f(x) = x^2$ Let $f : [0, 2] → \mathbb R$ be continuous and $f(2) = 0$. If $\lim \limits _{x \to 1} \frac {f(x) − 2} {\sqrt x − 1} = 1$, then prove that there exists $x \in [0, 2]$ such that $f(x) = x^2$.
I tried to use L'Hopital's rule to get $f'(1)=\frac 1 2$. But for the foolwing steps I have no idea. Please help!!

By the existence of the limit and $\sqrt x-1\to 0$ as $x\to 1$, we conclude that also $f(x)-2\to 0$ as $x\to 1$. Hence by continuity of $f$, $f(1)=2$. But then the IVT applied to $g(x):=f(x)-x^2$ on $[1,2]$ proves the claim (note that $g(1)=1>0$ and $g(2)=-4<0$).
