Limits at infinity of a function and its derivative

Question

I do not know how to solve this question, could anyone help me please?

Answer 1

The answer is A. Since $\lim_{x\to \infty}f'(x)$ exists, suppose by contradiction that $\lim_{x\to \infty}=L\ne 0.$ For some $r$ we have $$x>r\implies |f'(x)-L|<|L|/2\implies |f'(x)|>|L|/2\implies$$ $$\implies \exists y\in (x,x+1)\text { such that }|f(x+1)-f(x)|=\left|\frac {f(x+1)-f(x)}{(x+1)-x}\right|=$$ $$=|f'(y)|>|L|/2.$$ But if $L\ne 0$ then $\lim_{x\to \infty}f(x)$ cannot exist when $x>r\implies |f(x+1)-f(x)|>|L|/2.$

OR You can eliminate C,D, and E with $f(x)=\tan^{-1}x,$ and eliminate B on the grounds that there may be arbitrarily large $x$ where $f''(x)$ does not exist.

Answer 2

Hint:

Consider the map $f_1:(0,\infty)\to\Bbb R$ defined by $f_1(x)=1/x$. Options (d) and (e) can thus be eliminated.

Whereas the map $f_2:\Bbb R\to \Bbb R$ defined by $f(x)=1$ serves as a counter example for (c)

Answer 3

Hint: Use the mean value theorem on the interval $[n,n+1]$.