Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \cos x - x}}{{{x^2}}} = 1\]
Giải thích các bước giải:
\(\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = 1\)
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - \cos x - x}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {\sqrt {1 + 2x} - \left( {x + 1} \right)} \right] + \left( {1 - \cos x} \right)}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\left( {1 + 2x} \right) - {{\left( {x + 1} \right)}^2}}}{{\sqrt {1 + 2x} + x + 1}} + 1 - \sqrt {1 - {{\sin }^2}x} }}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{x^2}}}{{\sqrt {1 + 2x} + x + 1}} + \frac{{1 - 1 + {{\sin }^2}x}}{{1 + \sqrt {1 - {{\sin }^2}x} }}}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{{\sqrt {1 + 2x} + x + 1}} + \frac{{{{\sin }^2}x}}{{{x^2}.\left( {1 + \sqrt {1 - {{\sin }^2}x} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {1 + 2x} + x + 1}} + \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{1}{{1 + \sqrt {1 - {{\sin }^2}x} }}\\
= \frac{1}{{\sqrt {1 + 2.0} + 0 + 1}} + 1.1.\frac{1}{{1 + \sqrt {1 - {{\sin }^2}0} }}\\
= \frac{1}{2} + \frac{1}{2} = 1
\end{array}\)
