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Question Number 186267 by TUN last updated on 02/Feb/23

$$\underset{x\to 0}{\lim }\frac{{(1+2016x)}^{2017}-{(1+2017x)}^{2016}}{x^2}$$$$without{L}^{\prime }Horseries$$

Commented by JDamian last updated on 02/Feb/23

use binomial expansion

Answered by mahdipoor last updated on 02/Feb/23

$${(1+ax)}^b=1+b\left(ax\right)+\frac{b(b-1)}{2}{\left(ax\right)}^2$$$$+x^2(i_{3}x+i_4{x}^2+...+i_b{x}^{b-2})$$$$\Rightarrow A=\frac{{(1+2016x)}^{2017}-{(1+2017x)}^{2016}}{x^2}=$$$$\frac{2017\times 2016\times {2016}^2-2016\times 2015\times {2017}^2}{2}+$$$$(j_{3}x+...+j_{2017}{x}^{2015})$$$$limA,x\to 0=$$$$\frac{2017\times 2016\times {2016}^2-2016\times 2015\times {2017}^2}{2}$$$$\frac{2017\times 2016}{2}$$

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