6
$\begingroup$

enter image description here

How to do this question?

Tried separating the limit to

$$\lim_{x \rightarrow +\infty} f(x) + 2 \lim_{x \rightarrow +\infty} f'(x) + \lim_{x \rightarrow +\infty} f"(x) = k$$

but seems not working

$\endgroup$
4
  • $\begingroup$ Inequality $\lim_{x\to \infty} f(x)\neq 0$ suggests that limit exists. But if this limit exists, then limit of $f'(x)$ is zero and limit of second derivative is also zero. Then limit of function is $k$. $\endgroup$ Oct 5, 2022 at 10:38
  • 1
    $\begingroup$ @IvanKaznacheyeu $\lim_{+\infty}f$ exists $\not\Rightarrow\lim_{+\infty}f'$ exists. $\endgroup$ Oct 5, 2022 at 10:45
  • 3
    $\begingroup$ Let $\lim f(x)=A$, then $\lim (2f'(x)+f''(x))=k-A$, then $2f'(x)+f''(x)=k-A+o(1)$. Integrating gives $2f(x)+f'(x)=(k-A)x+o(x)$. Also $f(x)=A+o(1)=o(x)$, then $f'(x)=(2f(x)+f'(x))-2f(x)=(k-A)x+o(x)$. Integrating gives $f(x)=\frac12 (k-A)x^2+o(x^2)$. If $k\neq A$ then $\lim f(x)$ does not exist, then $k=A$. $\endgroup$ Oct 5, 2022 at 13:14
  • 1
    $\begingroup$ Please do not rely on pictures for the main parts of your post. Make the effort to type. $\endgroup$ Mar 29, 2023 at 13:41

2 Answers 2

26
$\begingroup$

Let $$F(x)=e^xf(x)\quad\text{and}\quad G(x)=e^x.$$The hypothesis rewrites$$\lim_{+\infty}\frac{F''}{G''}=k.$$By L'Hôpital's rule (used twice), this implies $$\lim_{+\infty}\frac{F'}{G'}=k\quad\text{and}\quad\lim_{+\infty}\frac FG=k,$$i.e. $\lim_{+\infty}(f+f')=k$ and $\lim_{+\infty}f=k$, whence the claim. Note that the hypothesis of non-nullity of $\lim_{+\infty}f$, nor even of its existence, was useless.

$\endgroup$
9
  • 2
    $\begingroup$ That's really cool! $\endgroup$
    – RT1
    Oct 4, 2022 at 12:44
  • $\begingroup$ But $\lim_{\infty} f(x)\neq 0$ for $\frac{\infty}{\infty}$. I saw this trick somewhere. $\endgroup$
    – Bob Dobbs
    Oct 4, 2022 at 13:23
  • $\begingroup$ $\infty\times 0=c$ isnt it? $\endgroup$
    – Bob Dobbs
    Oct 4, 2022 at 13:45
  • 1
    $\begingroup$ @AnneBauval Now I see the logic behind it, $F(x)/G(x)$ will be $f(x)$ and this would be general case. Your idea is brilliant. Thank you $\endgroup$
    – Magenta
    Oct 5, 2022 at 4:15
  • 1
    $\begingroup$ @Ivan Kaznacheyeu This hypothesis in L'Hospital's rule is in fact unnecessary. See e.g. the comments in en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_form $\endgroup$ Oct 5, 2022 at 12:49
1
$\begingroup$

We first prove this lemma:

Lemma: let h be a difrrentiable function in $(a,\infty)$ such that $\lim_{x \rightarrow +\infty} h(x) + h^{'}(x) =k$ thus $\lim_{x \rightarrow +\infty} h(x) = k$ and $\lim_{x \rightarrow +\infty} h^{'}(x) = 0$.

It's enough to prove that $lim_{x \to \infty}h(x)=k$.

Let $\epsilon > 0$ let's assume by contradiction that for all $N>0$ there exit $x>N$ such that $h(x) > k + \epsilon$. because $\lim_{x \to +\infty} h(x) + h^{'}(x) =k$ there exist $M>0$ such that for all $x>M$ we have $k-\frac{\epsilon}{2} < h(x)+h^{'}(x) < k-\frac{\epsilon}{2}$.

Notice that there exist $y>M$ such that $h(y) <k + \epsilon$ otherwise for all $y>M$ $h^{'}(y)\le k-\frac{\epsilon}{2} - h(y) \le k-\frac{\epsilon}{2} -(k + \epsilon)= -\frac{\epsilon}{2}$. and thus $\lim_{y \to \infty} f(y) = -\infty$ (from lagrange).

So let $y>M$ such that $h(y) < k+ \epsilon$ from the assumption there exist $x>y$ such that $h(x) > k +\epsilon$. From continouty of $h$ there exist $a \in (y,x)$ such that $h(a) = k + \epsilon$. Define $z = \sup\{b\in (y,x) | h(b) =k+ \epsilon\}$ notice that z is well defined as $a \in \{b\in (y,x) | h(b) =k+ \epsilon\}$. Also from continouty $h(z)=k + \epsilon$ and $z<x$. And finally notice that for every $c \in (z,x)$ $h(c) > k +\epsilon$(othewise there will be a point $p$ between $(z,x)$ such that $h(p)=k + \epsilon$ in cotrandiction to that that $z$ is the suprimum.

notice that $h(x)>h(z)$ and thus from lagrange there exist a point $b \in (z,x)$ such that $h'(b) > 0$. But this is a contradiction as for every point $c \in (z,x)$ $h^{'}(c)\le k-\frac{\epsilon}{2} - h(c) \le k-\frac{\epsilon}{2} -(k + \epsilon)= -\frac{\epsilon}{2}<0$

And thus there exist $N_1>0$ such that for every $x>N_1$ we have $h(x) \le k+ \epsilon$. In the same way we can prove that there exist $N_2>0$ such that for every $x>N_2$ we have $h(x) \ge k - \epsilon$. And thus for $N = \max\{N_1,N_2\}$ for all $x > N$ we have $|h(x) -k| \le \epsilon$.

And thus $lim_{x \to \infty}h(x)=k$. So we get the lemma.

Now define $h(x) = f(x) +f^{'}(x)$. Thus the assumtion in the question tells us that $\lim_{x \rightarrow +\infty} h(x) + h^{'}(x) =k$.And thus from the lemma we get that $\lim_{x \rightarrow +\infty} h(x) = k$ and $\lim_{x \rightarrow +\infty} h^{'}(x) = 0$.

Now $k = \lim_{x \rightarrow +\infty} h(x) = \lim_{x \rightarrow +\infty} f(x) + f'(x)$. Abd again from the lemma we get that $\lim_{x \rightarrow +\infty} f(x) = k$ and $\lim_{x \rightarrow +\infty} f^{'}(x) = 0$.

And now from limit arritmetic we can get that $\lim_{x \rightarrow +\infty} f^{''}(x) = \lim_{x \rightarrow +\infty} (f^{''}(x) +2f^{'}(x)+f(x))-2f^{'}(x)-f(x)=k+2\cdot 0 - k = 0$

$\endgroup$
2
  • 4
    $\begingroup$ This turned out to be very complicated, If someone has an idea how to make it more readable feel free to edit $\endgroup$
    – RT1
    Oct 4, 2022 at 11:41
  • $\begingroup$ I don't think that there is anybody that courageous... $\endgroup$
    – Bob Dobbs
    Oct 4, 2022 at 15:00

Not the answer you're looking for? Browse other questions tagged .