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Let $f(x)$ be differentiable on $(0,\infty)$ such that $\lim\limits_{x\to\infty}(f(x)+2024f'(x))=2025$. Show that $\lim\limits_{x\to\infty}f(x) = 2025$.

We can see that $f(x) + 2024f'(x) = 2024\left[\dfrac{1}{2024}f(x)+f'(x)\right]\implies\lim\limits_{x\to\infty}\left[\dfrac{1}{2024}f(x)+f'(x)\right] = \dfrac{2025}{2024}$.

Let $g(x) = e^{\frac{x}{2024}}f(x)$. Then $g(x)$ is differentiable on $(0,\infty)$, with $g'(x) = e^{\frac{x}{2024}}\left[\dfrac{1}{2024}f(x)+f'(x)\right]$. Hence, $\lim\limits_{x\to\infty}\dfrac{g'(x)}{e^{\frac{x}{2024}}}=\dfrac{2025}{2024}.$

We need to show that $\lim\limits_{x\to\infty}f(x) = \lim\limits_{x\to\infty}\dfrac{g(x)}{e^{\frac{x}{2024}}} = 2025.$ If I could show that $g(x)\to\infty$ as $x\to\infty$, then by using L'Hopital's Rule, I would be done. But I've been struggling to show that $g(x)\to\infty$. I'd greatly appreciate some help. Thank you!

ten_to_tenth
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    You should update this to $2024$ and $2025$. – Dietrich Burde Feb 20 '25 at 16:57
  • @DietrichBurde Okay, I've just updated it! :D – ten_to_tenth Feb 20 '25 at 17:02
  • @DietrichBurde Oops! I just rewound the clock. Thanks! :D – ten_to_tenth Feb 20 '25 at 17:08
  • You don't actually have to show what you're trying to show. It comes for free from the fact $e^{x/2024}$ tends to $\infty$, L'Hopital's rule, and the given limit, but if you really want to show it still.... .... For $x \geq 0$, $\frac{g'(x)}{e^{x/2024}} \leq g'(x)$. Do you know what the limit of the lower quantity tells us about the larger quantity? – Brian Moehring Feb 20 '25 at 17:14
  • @BrianMoehring Hmm the RHS behaves like a positive constant for large $x$, so if we integrate from a sufficiently large $M$ to $x$, then we would get $g(x) >$ a linear function, which implies that $g(x)$ tends to infinity. Is that what you mean? – ten_to_tenth Feb 20 '25 at 17:34
  • @BrianMoehring I'm recently made aware that L'Hopital's doesn't require that the numerator must tend to infinity, so I can apply the rule immediately to get the desired result. Is this what you mean by "It comes for free from..."?: https://math.stackexchange.com/questions/62916/how-to-show-that-lim-limits-x-to-infty-fx-0-implies-lim-limits-x?rq=1 – ten_to_tenth Feb 20 '25 at 17:40
  • One doesn't need to show that numerator tends to $\infty $ to apply L'Hospital's Rule. – Paramanand Singh Feb 20 '25 at 18:14

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Let $$ h(x)=\frac1{2024}f(x)+f'(x). $$ Since $$ \lim_{x\to\infty}h(x)=\frac{2025}{2024}>0 $$ there is $N\ge 2024$ such that $$ h(x)\ge 1, \ \forall x\ge N.$$ Now $$ g'(x)=e^{\frac x{2024}}h(x) $$ and hence $$ g(x)-g(N)=\int_N^xe^{\frac t{2024}}h(t)dt\ge x-N $$ which implies $$ \lim_{x\to\infty}g(x)=\infty. $$

xpaul
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