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

Answered by cortano1 last updated on 21/Oct/22

 With L′Hopital rule   L=lim_(x→−1)  ((2020x^(2019) −2021)/(3x^2 ))   =((−2020−2021)/3)= −((4041)/3)=−1347

WithLHopitalruleL=limx12020x201920213x2=202020213=40413=1347

Commented by mathlove last updated on 21/Oct/22

with out L′Hopital rule sir

withoutLHopitalrulesir

Answered by cortano1 last updated on 21/Oct/22

 let x+1=y   lim_(y→0)  (((y−1)^(2020) −2021(y−1)−2022)/((y−1)^3 +1))  = lim_(y→0)  (((1−y)^(2020) −2021y−1)/(y^3 −3y^2 +3y))  = lim_(y→0)  ((1−2020y−2021y−1)/(y(y^2 −3y+3)))  = lim_(y→0)  ((−4041y)/(y(y^2 −3y+3)))=−((4041)/3)

letx+1=ylimy0(y1)20202021(y1)2022(y1)3+1=limy0(1y)20202021y1y33y2+3y=limy012020y2021y1y(y23y+3)=limy04041yy(y23y+3)=40413

Commented by mathlove last updated on 22/Oct/22

thank sir

thanksir

Answered by kapoorshah last updated on 21/Oct/22

       lim_(x→ −1)  ((x^(2020)  − 1 − 2021x − 2021  )/((x + 1)( x^2  − x + 1)))        =lim_(x→ −1)  ((( x + 1)(x^(1019)  − x^(1018)  + x^(1017)  − x^(1016)  + ..... + x − 1) − 2021(x + 1)  )/((x + 1)( x^2  − x + 1)))         =lim_(x→ −1)  ((x^(1019)  − x^(1018)  + x^(1017)  − x^(1016)  + ..... + x − 1 − 2021)/( x^2  − x + 1))         = ((−1 − 1 − 1 − 1 − ..... − 1 − 1 − 2021)/( 1 + 1 + 1))         = ((− 2020 − 2021)/( 3))         = − 1347

limx1x202012021x2021(x+1)(x2x+1)=limx1(x+1)(x1019x1018+x1017x1016+.....+x1)2021(x+1)(x+1)(x2x+1)=limx1x1019x1018+x1017x1016+.....+x12021x2x+1=1111.....1120211+1+1=202020213=1347

Commented by mathlove last updated on 22/Oct/22

thanks sir

thankssir

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