题目

计算下列极限 (1) lim _(xarrow 0)((1-2{x)^2)}dfrac (1)(xsin x)-|||-__(1) lim _(xarrow 0)((1-2{x)^2)}dfrac (1)(xsin x)-|||-__(1) lim _(xarrow 0)((1-2{x)^2)}dfrac (1)(xsin x)-|||-__(1) lim _(xarrow 0)((1-2{x)^2)}dfrac (1)(xsin x)-|||-__

计算下列极限

$( 1 ) \lim \limits_{ x \to 0} ( 1 - 2 x ^ 2 ) ^ \frac{ 1} {x\sin x }$ $( 2 ) \lim\limits _{ x \to \infty }\Big ( \frac{ x ^ 2 + 2 } { x ^ 2 + 2 x + 1 }\Big ) ^ x$

$( 3 ) \lim\limits_ { n \to \infty } ( 1 + \frac{2 }{n} +\frac{ 2 }{ n ^ 2 } ) ^ n$

$( 4 ) \lim\limits _ { n \to \infty } \Big( \frac{ 2 n - 3 } { 2 n + 1 } \Big) ^ n$

题目解答

答案

解： $( 1 ) \lim \limits_{ x \to 0} ( 1 - 2 x ^ 2 ) ^ \frac{ 1} {x\sin x }= \lim \limits_{ x \to 0} ( 1 - 2 x ^ 2 ) ^{\frac{-1}{2x^2} \times\frac{ -2x^2} {x\sin x }}$

$=e^{\lim\limits_{x\to0}\frac{-2x}{\sin x}}=e^{-2}$

$( 2 ) \lim\limits _{ x \to \infty }\Big ( \frac{ x ^ 2 + 2 } { x ^ 2 + 2 x + 1 }\Big ) ^ x=\lim\limits _{ x \to \infty }\Big ( 1-\frac{ 2x-1 } { x ^ 2 + 2 x + 1 }\Big ) ^ x$

$=\lim\limits _{ x \to \infty }\Big ( 1-\frac{ 2x-1 } { x ^ 2 + 2 x + 1 }\Big ) ^ x=\lim\limits _{ x \to \infty }\Big (1- \frac{2x-1} { x ^ 2 + 2 x + 1 }\Big ) ^ {\frac{x^2+2x+1}{1-2x}\times\frac{x(1-2x)}{x^2+2x+1}}$

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

$( 3 ) \lim\limits_ { n \to \infty } ( 1 + \frac{2 }{n} +\frac{ 2 }{ n ^ 2 } ) ^ n=\lim\limits_{n\to\infty}(1+\frac{2n+2}{n^2})^n$

$=\lim\limits_{n\to\infty}(1+\frac{2n+2}{n^2})^{\frac{n^2}{2n+2}\times\frac{2n+2}{n}}=e^2$

$( 4 ) \lim\limits _ { n \to \infty } \Big( \frac{ 2 n - 3 } { 2 n + 1 } \Big) ^ n= \lim\limits _ { n \to \infty } \Big( 1-\frac{ 4 } { 2 n + 1 } \Big) ^ n$

$\lim\limits_{n\to\infty}(1-\frac{4}{2n+1})^{\frac{2n+1}{-4}\times\frac{-4n}{2n+1}}=e^{-2}$

解析

步骤 1：计算(1)的极限
$\lim _{x\rightarrow 0}(1-2{x}^{2})\dfrac {1}{x\sin x}$
$=\lim _{x\rightarrow 0}(1-2{x}^{2})$
$={e}^{x\rightarrow 0}\dfrac {-2x}{\sin x}$
$={e}^{-2}$
步骤 2：计算(2)的极限
$\lim _{x\rightarrow \infty }{(\dfrac {{x}^{2}+2}{{x}^{2}+2x+1})}^{x}$
$=\lim _{x\rightarrow \infty }(1-\dfrac {2x-1}{{x}^{2}+2x+1})^{x}$
$=\lim _{x\rightarrow +\infty }(1-\dfrac {2x-1}{{x}^{2}+2x+1})^{x}$
$={e}^{x+\infty }\dfrac {\alpha (1-2x)}{{x}^{2}+2x+1}$
$={e}^{-2}$
步骤 3：计算(3)的极限
$\lim _{n\rightarrow \infty }{(1+\dfrac {2}{n}+\dfrac {2}{{n}^{2}})}^{n}$
$=\lim _{n\rightarrow \infty }{(1+\dfrac {2n+2}{{n}^{2}})}^{n}$
$=\lim _{n\rightarrow \infty }{(1+\dfrac {2n+2}{{n}^{2}})}^{\dfrac {{n}^{2}}{2n+2}}\times \dfrac {2n+2}{n}$
$={e}^{2}$
步骤 4：计算(4)的极限
$\lim _{n\rightarrow \infty }{(\dfrac {2n-3}{2n+1})}^{n}$
$=\lim _{n\rightarrow \infty }{(1-\dfrac {4}{2n+1})}^{n}$
$=\lim _{n\rightarrow \infty }{(1-\dfrac {4}{2n+1})}^{\dfrac {2n+1}{-4}}\times \dfrac {-4n}{2n+1}$
$={e}^{-2}$