Limit of derivative function at infinity 
$f$ is a differentiable function on real line such that $$\lim_{x \to \infty}f(x)=1$$ and $$\lim_{x \to \infty}f'(x)=\alpha$$. Then what can be said about $\alpha$

*

*$\alpha=0$

*$\alpha$ may not be $0$ but $|\alpha| \le1$

*$\alpha\geq1$

*$\alpha\leq-1$

I cannot think of any viable option other than 1,but i cannot prove it. Any hint or solution would be appreciated!
 A: By the process of elimination:


*

*Taking $f(x)=1$ eliminates answers $3$ and $4$.

*Let's assume that $\alpha$ could be nonzero, that is, there exists some $f$ that satisfies both conditions, and $\alpha\neq 0$. Let $\lambda>\neq0$. Then, define $$g(x)=\lambda\cdot f(x) - \lambda + 1.$$ It is then easy to see that $g$ satisfies condition $1$, i.e. it has a limit of $1$, and its derivative has a limit of $\lambda$. Therefore, we have just proven that if some nonzero $\alpha$ is possible, then all real values are possible, so $|\alpha|\leq 1$ is not the correct answer either.


Therefore, $\alpha=0$ must be the correct answer.

Now I know, that's not what we really want here. We'd rather actually prove that $\alpha=0$. To do that, consider the following facts:


*

*If $\alpha > 0$, then there exists some $M$ such that $f'(x)>\frac\alpha2$ for all $x>M$.

*$f(x)=\int_0^x f'(t) dt + f(0)$

*For any integrable $g$, we have $\int_0^x g(t) dt=\int_0^M g(t)dt + \int_M^x g(t) dt$.

*For any pair of integrable functions $f, g$, if $g(x)>h(x)$ for all $x\in [a,b]$ then $\int_a^b g(x)dx > \int_a^b h(x)dx$.

A: For sufficiently large $ x $ , $ f'$ is locally integrable so we can apply FTC . Let $ a> 0$, By FTC $\lim_{x \to \infty} \int_{x-a}^{x+a} f'(y) dy=\lim_{x \to \infty}(f (x+a)-f (x-a))=0$
If you take $|\alpha| >0$ you get contradiction
A: Because $\lim_{x \to \infty}f(x)=1$, $y=1$ is a horizontal asymptote of the graph of $f(x)$.
Therefore, $\lim_{x \to \infty}f'(x)=0$.
