Is this an 'acceptable proof'?

Question

Let $f$ be a function such that$f\in C^1$ and $-2<f'(x)<-1 \forall x \in \mathbb{R}$ , prove that $lim_{x\rightarrow +\infty}f(x)=-\infty$

Using MVT on an Interval $I=[0,x]$, $x>0$

$\frac{f(x)-f(0)}{x}=f'(c)$ for some $c\in I$

then multiplying by $x$ and going through the limit as $x\rightarrow \infty$.

However I'm doubtful of the usage of the MVT on an interval like $I$.

My interpretation is that each time we want to 'approach' infinity we increase $x$ and we find another $c$ that satisfies the MVT. However, I'm doubtful we can use that as an arguement.

I know it can be proven using comparaison by integrals. Just want to make sure whether this is an acceptable proof or not.

Answer 1

In Mean value theorem, choice of $c$ depends on $x$. So replace $c$ by $c_x$ to make it precise.

Now by MVT, for any $x>0$

$f(x)=f(0)+xf'(c_x) < f(0)-x$

taking $x\rightarrow +\infty$ gives desired conclusion.

Answer 2

You may use MVT to show your claim via contradiction:

• For all $x\in \mathbb{R}$ you have $f'(x) < -1 \Rightarrow f$ strictly decreasing.
• Assume $-\infty < M = lim_{x\to +\infty}f(x)$.
• Choose $a \in \mathbb{R}$ with $M < f(a) < M+1$.
• Then, for any $x > a$ you have $M < f(x) < f(a) < M+1$

It follows for any $x > a+1$

$$|f'(\xi_x)|= \frac{|f(x) - f(a)|}{|x-a|} = \frac{f(a) - f(x)}{x-a} < \frac{M+1 - M}{1} = 1 \mbox{ Contradiction to } f'(x) < -1 $$