Khai triển Maclaurin của cosx đến x⁴

Ta có:
\(I = \int {\frac{{2dx}}{{\sqrt {{x^2} + 4x + 5} }}} = 2\int {\frac{{dx}}{{\sqrt {{{(x + 2)}^2} + 1} }}} \)
Đặt \(x + 2 = \tan t \Rightarrow dx = \frac{{dt}}{{{{\cos }^2}t}}\). Khi đó:
\(I = 2\int {\frac{{\frac{{dt}}{{{{\cos }^2}t}}}}{{\sqrt {{\tan }^2t + 1} }}} = 2\int {\frac{{\frac{{dt}}{{{{\cos }^2}t}}}}{{\frac{1}{{\cos t}}}}} = 2\int {\frac{{dt}}{{\cos t}}} = 2\int {\frac{{\cos tdt}}{{{{\cos }^2}t}}} = 2\int {\frac{{\cos tdt}}{{1 - {{\sin }^2}t}}} \)
\( = \int {\left( {\frac{{\cos t}}{{1 - \sin t}} + \frac{{\cos t}}{{1 + \sin t}}} \right)dt} = - 2\ln \left| {1 - \sin t} \right| + 2\ln \left| {1 + \sin t} \right| + C = 2\ln \left| {\frac{{1 + \sin t}}{{1 - \sin t}}} \right| + C\)
\( = 2\ln \left| {\frac{{1 + \frac{{x + 2}}{{\sqrt {{{(x + 2)}^2} + 1} }}}{{1 - \frac{{x + 2}}{{\sqrt {{{(x + 2)}^2} + 1} }}}}} \right| + C = 2\ln \left| {\frac{{\sqrt {{{(x + 2)}^2} + 1} + x + 2}}{{\sqrt {{{(x + 2)}^2} + 1} - x - 2}}} \right| + C\)
\( = 2\ln \left| {x + 2 + \sqrt {{x^2} + 4x + 5} } \right| + C\)

Để tính giới hạn này, ta chia cả tử và mẫu cho \({5^n}\):

\(\mathop {\lim }\limits_{x \to \infty } \frac{{{{5.2}^n} - {{3.5}^{n + 1}}}}{{{{100.2}^n} + {{2.5}^n}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{5{{\left( {\frac{2}{5}} \right)}^n} - 15}}{{{{100{{\left( {\frac{2}{5}} \right)}^n} + 2}}}\)

Vì \(\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{5}} \right)^n} = 0\) nên:

\(\mathop {\lim }\limits_{x \to \infty } \frac{{5{{\left( {\frac{2}{5}} \right)}^n} - 15}}{{{{100{{\left( {\frac{2}{5}} \right)}^n} + 2}}} = \frac{{0 - 15}}{{0 + 2}} = - \frac{{15}}{2}\)

Vậy, đáp án đúng là \(-\frac{{15}}{2}\).

Ta có: \({f'_ + }(0) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{\frac{1}{x}}}}}{x}\)

Đặt \(t = \frac{1}{x}\). Khi \(x \to {0^ + }\) thì \(t \to + \infty \).

Khi đó, \({f'_ + }(0) = \mathop {\lim }\limits_{t \to + \infty } t.{e^t} = + \infty \).