Is my proof correct?
If $\quad$ $\exists \ f''$ $\quad$ $f$ $\cup $ , $\quad f:\mathbb{R}\rightarrow\mathbb{R}$ ,$\quad$ $f$ $\\$ has two oblique asymptotes $a_1(x)=α_1x+b_1$ αnd $α_2(x)=α_2x+b_2$ $\\$ at $-\infty$ and$+\infty$ respectively.
$P(c,d)$ denote the points such that $\ $ $\ $ $d>a_1(c)$ ,$\ $ $d>a_2(c)$, $\ $ $d<f(c)$
Then:
Every $P(c,d)$ lies on exactly two tangent lines of $C_f$ $\iff \lim\limits_{x \to -\infty} x\Big(a_1-f'(x) \Big)=\lim\limits_{x \to +\infty}x\Big(a_2-f'(x)\Big)=0$
Proof
$(\Longleftarrow)$
By the definition of asymptotes: $\lim\limits_{x \to -\infty}\Big[f(x)-a_2x-b_2\Big]=0 $ $\lim\limits_{x \to -\infty} {f(x)\over x}=a_1$
Knowing $\lim\limits_{x \to -\infty}f'(x)$ $\ $exists $\Big( $ because $f'$ is strictly monotone $\Big)$ and $\lim\limits_{x \to -\infty}f(x)=\pm \infty $ depending on $α_1$, we apply L'Hôpital's rule as follows:
- $\lim\limits_{x \to -\infty}f'(x)=\lim\limits_{x \to -\infty}{ \big(f(x)\big) '\over \big(x \big)'}\overset{DLH}{=} \lim\limits_{x \to -\infty} {f(x)\over x}= a_1 $
$\lim\limits_{x \to -\infty}x(\alpha_1 -f'(x))=0\Longrightarrow\Big[ \lim\limits_{x \to -\infty} \alpha_1 x -xf'(x)\Big]=0\Longrightarrow \lim\limits_{x \to -\infty} \Big[\alpha_1 x-f'(x)+f(x)+xf'(x)\Big]=0 \overset{\alpha_1x-f(x)\to -b_1}{\Longrightarrow} \lim\limits_{x \to -\infty}\Big[f(x)- x f'(x) \Big]=b_1 $
- $\lim\limits_{x \to +\infty}\Big[x\Big(a_2-f'(x)\Big)\Big]=0\overset{\dots} {\Longrightarrow}\lim\limits_{x \to +\infty}\Big[f(x)- x f'(x) \Big]=b_2$
Therefore
$α_1(x)=x\lim\limits_{x \to -\infty}\Big[f'(x) \Big]+ \lim\limits_{x \to -\infty}\Big[f(x)- x f'(x) \Big]$
$α_2(x)=x\lim\limits_{x \to +\infty}\Big[f'(x) \Big]+ \lim\limits_{x \to +\infty}\Big[f(x)- x f'(x) \Big]$
The number of tangent lines of $C_f$ on which $P(c,d)$ lies, is given by the number of solutions to the equation $d-f(x)=f'(x)(c-x)$, which is equal to the number of roots of $g$, where $g(x)=f(x)+f'(x)(c-x)-d, \forall x\in \mathbb{R}$
$g'(x)=(a-x)f''(x)$ $\\$ , but $f''(x)>0$ ($f$ is convex).
So, g is strictly increasing for all $x\in (-\infty,c]$ and strictly decreasing for all $x\in [c,+\infty)$.
$P(c,d)$ lies above both asympotes and beneath $C_f$ , so we have the system of inequalities:
- $f(c)>d \quad (3) $
- $<d\Longrightarrow \quad c\lim\limits_{x \to -\infty}\Big[f'(x) \Big]+ \lim\limits_{x \to -\infty}\Big[f(x)- x f'(x) \Big]\Longrightarrow \quad \lim\limits_{x \to -\infty}\Big[f(x)+(c-x)f'(x)-d\Big]<0 \Longrightarrow\quad \lim\limits_{x \to -\infty}g(x)<0 \quad (4)$
- $<d\Longrightarrow c\lim\limits_{x \to +\infty}\Big[f'(x) \Big]+ \lim\limits_{x \to +\infty}\Big[f(x)- x f'(x) \Big] \Longrightarrow \quad \lim\limits_{x \to +\infty}\Big[f(x)+(c-x)f'(x)-d\Big]<0 \Longrightarrow\quad \lim\limits_{x \to +\infty}g(x)<0 \quad (5)$
also $\ $ $g(c)=f(c)+(c-c)f'(c)-d= f(c)-d \overset{(3)}{\Longrightarrow} \quad g(c)>0\quad (6)$
Because $f'' $ exists,$\ $ $ f$ , $f' $ are continuous (as differentiable functions), $g$ is continuous (as operations between continous functions). So we can apply Bolzano's theorem for $g$ .
$(4),(6)\longrightarrow g(c)g(t)<0$ $\ $for a $t$ near $-\infty$. By Bolzano's theorem and because $g$ is strictly monotone in $(-\infty, c]$, g has exactly one root in $(-\infty,c)$
$(6)\longrightarrow g(c)\neq 0$
$(5),(6)\longrightarrow g(c)g(r)<0$ $\ $for a $t$ near $+\infty$ By Bolzano's theorem and because g is strictly monotone in $[c, +\infty)$ g has exactly one root in $[c, +\infty)$
Therefore , $g$ has exactly two roots. Which means that the number of tangent lines of $C_f$ on which $P(c,d)$ lies are two. What was to be shown.
$(\Longrightarrow)$
Let $h(x)=f(x)-xf'(x)$
$h'(x)=-xf''(x)\Longrightarrow h'(x) >0 , x<0$ $\Longrightarrow$ $h$ is strictly increasing in $\ $ $(-\infty,0)$ $\Longrightarrow \lim\limits_{x \to -\infty} h(x)\in\mathbb{\overset{-}{R}}\Longrightarrow \lim\limits_{x \to -\infty}\Big[f(x)- x f'(x) \Big]\in\mathbb{\overset{-}{R}}\quad (1)$$d>a_1(c)\\ \Longrightarrow d>\alpha_1 c +b_1\\ \Longrightarrow \alpha_1 c-d<-b_1\\ \overset{\dots}{\Longrightarrow}\sup \Big(\alpha_1 c-d \Big)=-b_1\\ \Longrightarrow \sup \Big( \lim\limits_{x \to -\infty}\big[f(x)-xf'(x)\big]+a_1 c-d \Big)=\lim\limits_{x \to -\infty}\big[f(x)-xf'(x)\big]-b_1\\ \Longrightarrow \sup \Big( \lim\limits_{x \to -\infty}\big[f(x)-xf'(x) \big]+\lim\limits_{x \to -\infty}\big[ cf'(x)-d\big] \Big)=\lim\limits_{x \to -\infty}\big[f(x)-xf'(x)\big]-\lim\limits_{x \to -\infty}\big[f(x)-\alpha_1 x \big]\\ \Longrightarrow \sup \Big( \lim\limits_{x \to -\infty}\big[f(x)+f'(x)(c-x)-d\big]\Big)=\lim\limits_{x \to -\infty} x\Big(a_1-f'(x) \Big)\\ \Longrightarrow \sup \Big( \lim\limits_{x \to -\infty}\big[ g(x)\big] \Big)= \lim\limits_{x \to -\infty} x\Big(a_1-f'(x) \Big)\quad (2)$
But we want $g$ to have a root in $(-\infty, c)$, but $g(c)>0$, so it must be: $\lim\limits_{x \to -\infty}\big[ g(x)\big]<0$ and therefore : $0\geq \sup \Big( \lim\limits_{x \to -\infty}\big[ g(x)\big]\Big) \quad (3)$
$(2),(3)\Longrightarrow 0\geq \lim\limits_{x \to -\infty} x\Big(a_1-f'(x) \Big)\quad (4)$
- $f \quad \cup\Longrightarrow f'(x)>\alpha_1,\forall x\in\mathbb{R}\Longrightarrow \lim\limits_{x \to -\infty} x\Big(a_1-f'(x)\Big) \geq0\quad(5)$
$(4),(5)\Longrightarrow \lim\limits_{x \to -\infty} x\Big(a_1-f'(x) \Big)=0$
Similarly we can show $\lim\limits_{x \to +\infty} x\Big(a_2-f'(x) \Big)=0$
$\square$
