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Question Number 178751 by mathlove last updated on 21/Oct/22

Answered by cortano1 last updated on 21/Oct/22

$$\mathrm{With}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{rule}$$$$\mathrm{L}=\underset{x\to -1}{\lim }\frac{{2020\mathrm{x}}^{2019}-2021}{{3\mathrm{x}}^2}$$$$=\frac{-2020-2021}{3}=-\frac{4041}{3}=-1347$$

Commented by mathlove last updated on 21/Oct/22

$$without{L}^{\prime }Hopitalrulesir$$

Answered by cortano1 last updated on 21/Oct/22

$$\mathrm{let}\mathrm{x}+1=\mathrm{y}$$$$\underset{\mathrm{y}\to 0}{\lim }\frac{{(\mathrm{y}-1)}^{2020}-2021(\mathrm{y}-1)-2022}{{(\mathrm{y}-1)}^3+1}$$$$=\underset{\mathrm{y}\to 0}{\lim }\frac{{(1-\mathrm{y})}^{2020}-2021\mathrm{y}-1}{{\mathrm{y}}^3-{3\mathrm{y}}^2+3\mathrm{y}}$$$$=\underset{\mathrm{y}\to 0}{\lim }\frac{1-2020\mathrm{y}-2021\mathrm{y}-1}{\mathrm{y}({\mathrm{y}}^2-3\mathrm{y}+3)}$$$$=\underset{\mathrm{y}\to 0}{\lim }\frac{-4041\mathrm{y}}{\mathrm{y}({\mathrm{y}}^2-3\mathrm{y}+3)}=-\frac{4041}{3}$$$$$$

Commented by mathlove last updated on 22/Oct/22

$$thanksir$$

Answered by kapoorshah last updated on 21/Oct/22

$$\underset{x\to -1}{\lim }\frac{x^{2020}-1-2021x-2021}{(x+1)(x^2-x+1)}$$$$=\underset{x\to -1}{\lim }\frac{(x+1)(x^{1019}-x^{1018}+x^{1017}-x^{1016}+.....+x-1)-2021(x+1)}{(x+1)(x^2-x+1)}$$$$=\underset{x\to -1}{\lim }\frac{x^{1019}-x^{1018}+x^{1017}-x^{1016}+.....+x-1-2021}{x^2-x+1}$$$$=\frac{-1-1-1-1-.....-1-1-2021}{1+1+1}$$$$=\frac{-2020-2021}{3}$$$$=-1347$$$$$$$$$$$$$$

Commented by mathlove last updated on 22/Oct/22

$$thankssir$$

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