# If $\lim_{x \to \infty}f'(x)=+\infty$ then $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and $\lim_{x \to \infty}f(x)=+\infty$.

Let $f$ be differentiable and let $\lim_{x \to \infty}f'(x)=+\infty$ prove that: 1) $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and 2) $\lim_{x \to \infty}f(x)=+\infty$.

1) I’ll prove by contradiction. Let $\lim_{x \to \infty}(f(x)-f(x-1))=a$, where $a$ is from and $\Bbb R$. $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a$, where c is from $(x-1,x)$ (mean value theorem). As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a$. Which is a contradictition since we have that $\lim_{x \to \infty}f'(x)=+\infty$ .

2) I will also prove this by contradiction. We have that $\lim_{x \to \infty}f(x)=a$ This is equivalent to $\lim_{x \to \infty}f(x)-f(x-1)=0$. I use mean value theorem and get $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0$, where c is from $(x-1,x)$. As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0$. Which is a contractition since we have that $\lim_{x \to \infty}f'(x)=+\infty$ .

#### Solutions Collecting From Web of "If $\lim_{x \to \infty}f'(x)=+\infty$ then $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and $\lim_{x \to \infty}f(x)=+\infty$."

There is no need to prove the statement by contradiction (also notice the issues that the other users pointed out). The fact that:
$$\lim_{x\to +\infty}f'(x) = +\infty$$
implies that for any $x> x_0$, $f'(x)\geq N.$ Since by the Lagrange theorem:
$$f(x)-f(x-1) = f'(\xi),\qquad \xi\in[x-1,x]$$
for any $x>x_0 + 1$ we have:
$$f(x)-f(x-1) \geq N,$$
but since $N$ is an arbitrarily big number,
$$\lim_{x\to +\infty} f(x)-f(x-1) = +\infty.$$
The Lagrange theorem also implies:
$$f(x_0+y)-f(x_0) = y\cdot f'(\xi) \geq Ny, \qquad \xi\in[x_0,x_0+y],$$
hence
$$f(x_0+y) \geq f(x_0) + Ny$$
gives:
$$\lim_{x\to +\infty} f(x) = +\infty.$$

1) is wrong because the converse of having an infinite limit is not just the existence of a finite limit. You could also have no limit at all.

2) Same problem.