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This question came in ISI entrance exam 2022 in India. So it was for high school students, so I am not expecting much use of real analysis. Tried with the well-known limit of $e$ but no luck. I am not getting how to proceed. Also does it depend on the degree of the polynomial or can we generalize it? Any help will be truly appreciated.

Let $$ P(x)=1+2 x+7 x^{2}+13 x^{3}, x \in \mathbb{R} . $$ Calculate for all $x \in \mathbb{R}$, $$ \lim _{n \rightarrow \infty}\left(P\left(\frac{x}{n}\right)\right)^{n} $$

Sayantan
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1 Answers1

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I believe a generalization is possible, mimicking this series expansion approach. First, take the natural log of the limit, and using the algebra of limits you get $$\ln\left(\lim_{n\to \infty} \left(P\left(\frac{x}{n}\right)\right)^n\right) = \lim_{n \to \infty} n\ln\left(P\left(\frac{x}{n}\right)\right)$$ Expand this into $$\lim_{n \to \infty} n[(P\left(\frac{x}{n}\right)-1) + O((P\left(\frac{x}{n}\right)-1)^2)] = 2x+O(\frac{1}{n})$$ Hence the answer $e^{2x}$.

geoalg
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  • Great answer! It would be helpful to include a comment about why the power series of $\log$ is valid here - which is that for sufficiently large $n$, $x/n$ is close enough to $0$ that $0 < P(x/n) < 2,$ so the power series expansion converges. – ABanerjee May 10 '22 at 06:41
  • @geoalg but I don't think this 0<(/)<2, is holding here, so will the expansion work? – Sayantan May 10 '22 at 08:10
  • @Sayantan It holds for fixed $x$ and large enough $n$. We just ignore the smaller $n$ since we are taking the limit as $n\to\infty$. (Note that $P(x/n)\to1$). – Milten May 10 '22 at 11:48