-Kmath科数- 中考、高考、考验数学题库

设$ \lim \limits _{x \rightarrow \infty } \left ( \dfrac {x+c}{x-c} \right )^{ \dfrac {2x^{2}+3x+2}{x+1}}= \lim \limits _{x \rightarrow 0}( \cos x+x^{2})^{ \dfrac {1}{ \sin ^{2}x}}$, 则$c=\underline{\quad\quad\quad}$.

0 人点赞

6 次查看

我来讲解

答案：

$\dfrac {1}{8}$

解析：

$\lim _{x \rightarrow \infty}\left(\dfrac{x+c}{x-c}\right)^{\dfrac{2 x^{2}+3 x+2}{x+1}}=\lim _{x \rightarrow \infty}\left[\left(1+\dfrac{2 c}{x-c}\right)^{\dfrac{x-c}{2 c}}\right]^{\dfrac{2 x^{2}+3 x+2}{x+1} \cdot \dfrac{2 c}{x-c}}$

$={e}^{\dfrac{2 c}{x \rightarrow \lim \frac{2 x^{2}+3 x+2}{x}(x+1)(x-c)}}=\mathrm{e}^{4 c}$,

$ \lim _{x \rightarrow 0}\left(\cos x+x^{2}\right)^{\dfrac{1}{\operatorname{sn}^{2} x}}=\lim _{x \rightarrow 0}\left\{\left[1+\left(\cos x-1+x^{2}\right)\right]^{\dfrac{1}{\cos x-1+x^{2}}}\right\}^{\dfrac{\cos x-1+x^{2}}{\sin ^{2} x}}$

$=\mathrm{e}^{\lim \dfrac{\operatorname{los} x-1+x^{2}}{\sin ^{2} x}} \\ =\mathrm{e}^{\lim _{x \rightarrow 0} \dfrac{\cos x-1+x^{2}}{x^{2}}}=\mathrm{e}^{\lim _{x \rightarrow 0} \dfrac{\cos x-1}{x^{2}+1}}=\mathrm{e}^{\dfrac{1}{2}} $

得 $c=\frac{1}{8} .$

①点击首页查看更多试卷和试题 , 点击查看本题所在试卷
② 下载本题Word版或下载本题PDF版点击赞助本站

关闭