bai tap gioi han kho can giai gap

N

nguyenbahiep1

đáp án bài này là 2012 + 2011..........................................................................................................
 
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N

noinhobinhyen

Đặt $t=x-2011 ; x \to 2011 \Rightarrow t \to 0$

$I=lim_{t \to 0} \dfrac{(t+1)^{2012}-(1-t)^{2011}}{t}$

$=lim_{t \to 0} \dfrac{(t+1)^{2012}}{t}-lim_{t \to 0} \dfrac{(1-t)^{2011}}{t}$

Ta có :

$A = lim_{t \to 0} \dfrac{(t+1)^{2012}}{t} = 2012+lim_{t \to 0} \dfrac{1}{t}$

$B= lim_{t \to 0} \dfrac{(1-t)^{2011}}{t} = -2011+lim_{t \to 0} \dfrac{1}{t}$

$\Rightarrow I=A-B=4023$

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mình xin giải thích vì sao lại có $A=2012+lim_{t \to 0} \dfrac{1}{t}$

thật vậy ta xét khai triển $(t+1)^{2012} = 1+a_1t+a_2t^2+...+a_{2012}t^{2012}$

$\Rightarrow A=\lim_{t \to 0} [\dfrac{1}{t}+a_1+a_2t+a_3t^2+...+a_{2012}t^{2011}$

$=[\lim_{t \to 0} \dfrac{1}{t}]+a_1 = 2012+lim_{t \to 0} \dfrac{1}{t}$
 
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