Tính lim x đến + vô cùng căn bậc hai của x^2 + 2/x + 1

Lời giải:

Ta có \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 2} }}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2}\left( {1 + \frac{2}{{{x^2}}}} \right)} }}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{2}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{2}{{{x^2}}}} }}{{1 + \frac{1}{x}}}\)

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