Let $f(x)$ be a differentiable function s. t. $\lim\limits_{x\to\infty}f^\prime (x)=1$. Prove that $\lim\limits_{x\to\infty} f(x)=\infty$ and $f(x)$ is asymptotic to the identity function $h(x)=x$.
I thought in define $f_{n}(x)=f(x+n)$ and consider $\displaystyle\int \lim_{n\to\infty}f_{n}(x) dx$ in order to pass the limit inside the integral and have something like this: $\displaystyle\int dx=\int\lim_{x\to\infty}f^{\prime}(x)dx=\int\lim_{n\to\infty}f^{\prime}(x+n)$, but for this I must define a lower and upper limit for the integral.
I also thought in proving that $\displaystyle\lim_{x\to\infty}\Big|\frac{f(x)}{x}\Big|=1$, and for this I thought in tangent line approximation of $f$, i.e, if f is differentiable at $x=a$, then for $x$ close to $a$ we have that $f(x)\approx f(a)+f^{\prime}(a)(x-a)$, but I don´t know how to continue.
To prove that $\lim_{x\rightarrow+\infty}f(x)=+\infty$:
Since $\lim_{x\rightarrow+\infty}f'(x)=1$, there exists $x_{0}$ such that $f'(x)>\frac{1}{2}$ whenever $x\geq x_{0}$. By mean-value theorem, for any $x\in(x_{0},\infty)$, there exists $\xi_{x}\in(x_{0},x)$ such that $f(x)-f(x_{0})=f'(\xi_{x})(x-x_{0})$. Therefore, we have estimation: \begin{eqnarray*} f(x) & = & f(x_{0})+f'(\xi_{x})(x-x_{0})\\ & \geq & \frac{1}{2}(x-x_{0})+ f(x_0)\\ & \rightarrow & +\infty \end{eqnarray*} as $x\rightarrow+\infty$. This shows that $\lim_{x\rightarrow+\infty}f(x)=+\infty$.
To prove that $\lim_{x\rightarrow+\infty}\frac{f(x)}{x}=1$:
Observe that the limit is of $\frac{+\infty}{+\infty}$ form, so L'Hospital rule is applicable. We have that \begin{eqnarray*} & & \lim_{x\rightarrow+\infty}\frac{f(x)}{x}\\ & = & \lim_{x\rightarrow+\infty}\frac{f'(x)}{1}\\ & = & 1. \end{eqnarray*}