https://spherical-harmonics.hateblo.jp/entry/Todai/1950/Kaiseki_II

[3] 次の極限値を求む．

(1) $\displaystyle\lim_{x\to 1}\dfrac{x^2-3x+2}{x^2-4x+3}$

(2) $\displaystyle\lim_{n\to\infty}\{\log n-\log(n-1)\}$

(3) $\displaystyle\lim_{x\to\infty}(\sqrt{x+1}-\sqrt{x-1})$

(4) $\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos 3\theta}{\theta^2}$

2024.11.04記

[解答]
(1) $\displaystyle\lim_{x\to 1}\dfrac{x^2-3x+2}{x^2-4x+3}$ $=\displaystyle\lim_{x\to 1}\dfrac{x-2}{x-3}=\dfrac{1}{2}$ ．

(2) $\displaystyle\lim_{n\to\infty}\{\log n-\log(n-1)\}$ $=\displaystyle\lim_{n\to\infty}\log \left(1+\dfrac{1}{n-1}\right)=\log 1=0$ ．

(3) $\displaystyle\lim_{x\to\infty}(\sqrt{x+1}-\sqrt{x-1})$ $=\displaystyle\lim_{x\to\infty}\dfrac{2}{\sqrt{x+1}+\sqrt{x-1}}=0$ ．

(4) $\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos 3\theta}{\theta^2}$ $=\displaystyle\lim_{\theta\to 0}\dfrac{1-\cos 3\theta}{(3\theta)^2}\cdot 9=\dfrac{9}{2}$ ．