the choice of 2 when proving the limit when $x\to\pm\infty$ Suppose that $f$ is a continuous function on $\mathbb{R}$ and $\lim_{x\to -\infty}f(x)$ and $\lim_{x\to -\infty}f'(x)$ exist. 
Show that $\lim_{x\to -\infty}f'(x)=0$ 
A common way to show this is proving by contradiction.(See https://www.physicsforums.com/threads/limit-of-derivative-as-x-goes-to-infinity.502972/) 
So if $L>0$, then there is some $N$ such that $f'(x)>L/2$ for all $x>N$, then when we integrate we will see the contradiction. Here, I don't quite see the reason for "2" from the denominator? Any help is appreciated!
Could anyone explain to me what makes the choice $L/2$? my background in real analysis is quite weak here.
 A: The proof works with any denominator because $L\gt 0$ yields $\frac{L}{n}\gt 0$ Chosing $2$ is just picking one that works
A: It is more convenient to handle the case when $x \to \infty$ and to treat the original question of $x \to -\infty$ by considering the function $f(-x)$ when $x \to \infty$.
Assuming that $\lim_{x \to \infty}f'(x) = L > 0$ means that $f'(x)$ tends to a positive limit as $x \to \infty$. This means that the function $f'(x)$ is ultimately positive as $x \to \infty$. A formal proof of the previous statement requires the use of $\epsilon,\delta$ argument. Thus for any given $\epsilon > 0$ we have an $M > 0$ such that $$L - \epsilon < f'(x) < L + \epsilon$$ for $x > M$.
Now to make sure that $f'(x) > 0$ for $x > M$ (this is the meaning of $f'(x)$ is ultimately positive), it is just sufficient to choose $\epsilon$ such that $L - \epsilon \geq 0$ i.e. $0 < \epsilon \leq L$ and this will give $$0 \leq L - \epsilon < f'(x) < L + \epsilon$$ However note that if a function $g(x)$ is ultimately positive for all large $x$ then it does not mean that its limit is positive. For example $g(x) = 1/x$ and $g(x) > 0$ for all $x > 0$ yet its limit as $x \to \infty$ is $0$.
Thus for a function like $f'(x)$ with positive limit $L$ there has to be something extra compared to the functions like $g(x) = 1/x$ above. A positive limit ensures that the function $f'(x)$ is ultimately positive and a bit away from $0$. Thus for our $f'(x)$ we have a number $k > 0$ such that $0 < k < f'(x)$ for all large $x$. In order to achieve this we just need to choose $\epsilon$ such that $k = L - \epsilon > 0$ (earlier we just wanted $L - \epsilon \geq 0$). Thus any $\epsilon$ with $0 < \epsilon < L$ is sufficient to show that for large $x$ the values of $f'(x)$ are away from $0$ by a margin $k = L - \epsilon$. This property is much stronger than saying that $f'(x)$ is ultimately positive.
Now what do we gain by this extra property of $f'(x)$? Well for that we need to consider the function $h(x) = f(x) - kx$ so that $h'(x) = f'(x) - k > 0$ for all for large $x$ (i.e. $x > M$ for some positive $M$). This means that $h(x)$ is strictly increasing in the interval $[M, \infty)$. So $h(x) > h(M)$ for all $x > M$. Therefore $f(x) - kx > f(M) - kM$ for all $x > M$. This further means that $$f(x) > kx + f(M) - kM$$ Now we can see that in the above equation the RHS tends to $\infty$ as $x \to \infty$ whereas the LHS $f(x)$ has a finite limit. Thus we get a contradiction.
A similar contradiction is obtained if $L < 0$. Thus the only option is to have $L = 0$. One can clearly see that the choice $\epsilon = L$ does give us that $f'(x)$ is ultimately positive but it does not give the extra feature that $f'(x)$ is away from $0$ by a fixed margin which is the key to obtaining a contradiction. Hence we have to avoid $\epsilon = L$.
A: I wish to show the contradiction for the case $\lambda<0$. So any correction is appreciated!
Suppose $L>0$, then there exists $N>0$ such that
$$
f'(x)<L/2\quad\forall\quad x\le -N
$$
Therefore, for $x<-N$, we have $f(-N)-f(x)=\int_{x}^{-N}f'(s)\,ds<\int_{x}^{-N}\frac{L}{2}(-N-x)$.
Therefore $f(x)>\frac{L}{2}N+\frac{L}{2}x+f(-N)\to +\infty\,\,\text{as}\,\,x\to-\infty$,
so $\lim_{x\to-\infty}f(x)=+\infty$, which is a contradiction with the fact $f(-\infty)=1$.
Combine with the answer from Mauro, $L=0$ holds. (Note, I used positive $N$ here).
A: The proof requires a number that is intermediate between $0$ and $L$. It cannot be $0$ otherwise the integral wouldn't diverge, and it must be smaller than $L$ to keep the accumulation point.
Instead of $\dfrac L2$, any $0<M<L$ could have been used.
