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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