# How to calculate infinite limit for function and its derivative? [closed]

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

• 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 $k$. Oct 5, 2022 at 10:38
• @IvanKaznacheyeu $\lim_{+\infty}f$ exists $\not\Rightarrow\lim_{+\infty}f'$ exists. Oct 5, 2022 at 10:45
• 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 $k=A$. Oct 5, 2022 at 13:14
• Please do not rely on pictures for the main parts of your post. Make the effort to type. Mar 29, 2023 at 13:41

## 2 Answers

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.

• That's really cool!
– RT1
Oct 4, 2022 at 12:44
• But $\lim_{\infty} f(x)\neq 0$ for $\frac{\infty}{\infty}$. I saw this trick somewhere. Oct 4, 2022 at 13:23
• $\infty\times 0=c$ isnt it? Oct 4, 2022 at 13:45
• @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 Oct 5, 2022 at 4:15
• @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 Oct 5, 2022 at 12:49

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 $$h(y) otherwise for all $$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 $$z. And finally notice that for every $$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 $$\lim_{x \rightarrow +\infty} f^{''}(x) = \lim_{x \rightarrow +\infty} (f^{''}(x) +2f^{'}(x)+f(x))-2f^{'}(x)-f(x)=k+2\cdot 0 - k = 0$$

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