夾擠原理題型 (4)

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Evaluate the limit or show that it does not exist.
\(\lim\limits_{x \to \infty } \left( {\frac{1}{{\sqrt {{x^2} + 1} }} + \frac{1}{{\sqrt {{x^2} + 2} }} + \frac{1}{{\sqrt {{x^2} + 3} }} + ... + \frac{1}{{\sqrt {{x^2} + x} }}} \right)\) ？【96交大財金所】

詳解：當 \(x > 0\)，由不等式

\(\begin{array}{l}\frac{x}{{\sqrt {{x^2} + x} }} = \frac{{\rm{1}}}{{\sqrt {{x^2} + x} }} + \frac{{\rm{1}}}{{\sqrt {{x^2} + x} }} + \cdots \cdots + \frac{{\rm{1}}}{{\sqrt {{x^2} + x} }}\\\quad \quad \quad \quad < \frac{{\rm{1}}}{{\sqrt {{x^2} + 1} }} + \frac{{\rm{1}}}{{\sqrt {{x^2} + 2} }} + \cdots \cdots + \frac{{\rm{1}}}{{\sqrt {{x^2} + x} }}\\\quad \quad \quad \quad < \frac{{\rm{1}}}{{\sqrt {{x^2} + 1} }} + \frac{{\rm{1}}}{{\sqrt {{x^2} + 1} }} + \cdots \cdots + \frac{{\rm{1}}}{{\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }}\end{array}\)

\( \Rightarrow \frac{x}{{\sqrt {{x^2} + x} }} < \frac{{\rm{1}}}{{\sqrt {{x^2} + 1} }} + \frac{{\rm{1}}}{{\sqrt {{x^2} + 2} }} + \cdots \cdots + \frac{{\rm{1}}}{{\sqrt {{x^2} + x} }} < \frac{x}{{\sqrt {{x^2} + 1} }}\)

\(\lim\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + x} }} = \lim\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + 1} }} = 1\)

根據夾擠原理得 \(\mathop {\lim }\limits_{x \to \infty } \frac{{\rm{1}}}{{\sqrt {{x^2} + {\rm{1}}} }}{\rm{ + }}\frac{{\rm{1}}}{{\sqrt {{x^2} + {\rm{2}}} }}{\rm{ + }} \cdots \cdots {\rm{ + }}\frac{{\rm{1}}}{{\sqrt {{x^2} + x} }}{\rm{ = 1}}\)

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