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I'm trying to solve this problem:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function so that $$\lim_{x\rightarrow +\infty}f(x)+ \frac{\ln(x)^2}{\ln(x)-1}f'(x)=0$$ Prove that $\lim_{x\rightarrow +\infty}f(x)$ exists and find it.

I've done it in the special case $f(x)+ \frac{\ln(x)^2}{\ln(x)-1}f'(x)=0$ for all $x$, solving the differential equation $f(x)=c_{1}e^{-\frac{x}{\ln(x}}$ and clearly $\lim_{x\rightarrow +\infty}f(x)=0$.

How can I proceed in the general case with the limit?

Xander Henderson
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Carloss
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2 Answers2

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Applying L'Hospital's theorem to $$ \lim_{x \to \infty} \frac{e^{x/\ln(x)} f(x)}{e^{x/\ln(x)}} $$ shows that $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( f(x)+ \frac{\ln(x)^2}{\ln(x)-1}f'(x)\right) $$ if the latter limit exists.

More generally, if $g: [x_0, \infty) \to \Bbb R$ is an increasing differentiable function with $\lim_{x \to \infty} g(x) = +\infty$ and $f: [x_0, \infty) \to \Bbb R$ is differentiable then $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \frac{f'(x)}{g'(x)} + f(x) \right) $$ if the latter limit exists.

This is a generalization of the approach used before in special cases, like Show that $\lim_{x \to +\infty}\left(f(x)+f'(x)\right)=0 \Rightarrow \lim_{x \to +\infty} f(x)=0$ and If $f(x)+\frac{f'(x)}{x}\to0$ then $f(x)\to0$.

Martin R
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Here is my approach, although I must admit it's not as slick as the approach developed by @Martin R

Set $p(x)=\frac{1}{\ln(x)}-\frac{1}{\ln^2(x)}$ and pick $\epsilon>0.$

Find $x_0>e$ so that, for any $x\geq x_0$, we have $$\Big|f(x)+\frac{\ln^2(x)}{\ln(x)-1}f'(x)\Big|<\epsilon$$ This means $$\forall x\geq x_0: -\epsilon p(x)e^{\int_{x_0}^xp(s)\mathrm{d}s}< \frac{d}{dx}\Bigg(e^{\int_{x_0}^xp(s)\mathrm{d}s}f(x)\Bigg)<\epsilon p(x)e^{\int_{x_0}^xp(s)\mathrm{d}s}$$ But this is the same as saying $$\forall x\geq x_0: -\epsilon\frac{d}{dx}\Big(e^{\int_{x_0}^xp(s)\mathrm{d}s}\Big)< \frac{d}{dx}\Bigg(e^{\int_{x_0}^xp(s)\mathrm{d}s}f(x)\Bigg)<\epsilon \frac{d}{dx}\Big(e^{\int_{x_0}^xp(s)\mathrm{d}s}\Big)$$ Integrate from $x_0$ to $x$ and get $$\forall x\geq x_0:-\epsilon\Big(e^{\int_{x_0}^xp(s)\mathrm{d}s}-1\Big)<e^{\int_{x_0}^xp(s)\mathrm{d}s}f(x)-f(x_0)<\epsilon\Big(e^{\int_{x_0}^xp(s)\mathrm{d}s}-1\Big)$$ Divide through by $e^{\int_{x_0}^xp(s)\mathrm{d}s}$. $$\forall x\geq x_0: -\epsilon\Big(1-e^{-\int_{x_0}^xp(s)\mathrm{d}s}\Big)<f(x)-f(x_0)e^{-\int_{x_0}^xp(s)\mathrm{d}s}<\epsilon\Big(1-e^{-\int_{x_0}^xp(s)\mathrm{d}s}\Big)$$ It's not hard to show that $\int_{x_0}^{\infty} p(s)\mathrm{d}s=\infty$ so take $x\rightarrow \infty$ and invoke squeeze theorem to see $\Big|\lim_{x\rightarrow \infty}f(x)\Big|<\epsilon$. But $\epsilon>0$ is arbitrary, revealing $f(x)\rightarrow 0$ as $x\rightarrow \infty$

Matthew H.
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