Let $f$ be a continuous and positive function on $\mathbb{R}_{+}$ such that $\displaystyle\underset{x \to \infty}{\lim} \frac{f(x)}{x} <1$.
Prove the equation $$f(x)=x$$ has at least one solution on $\mathbb{R}_{+}$.
Answer
If $f(0)=0$, done. If not $g(x)=f(x)-x, g(0)>0$ and there exists $x>0$ such that $f(x)/x<1$ which implies that $g(x)=f(x)-x<0$, applies IVT at $g$ in $[0,x]$.
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