Compute \lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}}.
Proposed by Arkady Alt.
Solution:
We have \lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}}= \dfrac{\lim_{n \to \infty} \left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\lim_{n \to \infty} \left(1+\frac{1}{n+b}\right)^{n^2}}.
Since \left(1+\frac{1}{n(n+a)}\right)^{n^3}=e^{n^3 \log \left(1+\frac{1}{n(n+a)}\right)} \sim e^n as n \to \infty and \left(1+\frac{1}{n+b}\right)^{n^2}=e^{n^2 \log \left(1+\frac{1}{n+b}\right)} \sim e^n as n \to \infty, we have
\lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}} \sim \dfrac{e^n}{e^n}=1.
Note: the official problem was modified lately.
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