题目
lim _(narrow infty )(dfrac (1)({n)^2+n+1}+dfrac (2)({n)^2+n+2}+... +dfrac (n)({n)^2+n+n}).
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题目解答
答案
解析:
对原式进行缩放,得:
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解析
步骤 1：缩放原式
对原式进行缩放，得到两个不等式：
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+n}+\dfrac {2}{{n}^{2}+n+n}+\cdots +\dfrac {n}{{n}^{2}+n+n})$$
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+2}+\cdots +\dfrac {n}{{n}^{2}+n+n})$$
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+1}+\cdots +\dfrac {n}{{n}^{2}+n+1})$$
步骤 2：计算不等式的上下界
计算不等式的上下界，得到：
$$\dfrac {1}{2}\leqslant \lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+2}+\cdots +\dfrac {n}{{n}^{2}+n+n})\leqslant \dfrac {1}{2}$$
步骤 3：应用夹逼准则
根据夹逼准则，原式等于 $\dfrac {1}{2}$。
对原式进行缩放，得到两个不等式：
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+n}+\dfrac {2}{{n}^{2}+n+n}+\cdots +\dfrac {n}{{n}^{2}+n+n})$$
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+2}+\cdots +\dfrac {n}{{n}^{2}+n+n})$$
$$\lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+1}+\cdots +\dfrac {n}{{n}^{2}+n+1})$$
步骤 2：计算不等式的上下界
计算不等式的上下界，得到：
$$\dfrac {1}{2}\leqslant \lim _{n\rightarrow \infty }(\dfrac {1}{{n}^{2}+n+1}+\dfrac {2}{{n}^{2}+n+2}+\cdots +\dfrac {n}{{n}^{2}+n+n})\leqslant \dfrac {1}{2}$$
步骤 3：应用夹逼准则
根据夹逼准则，原式等于 $\dfrac {1}{2}$。