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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

$$\:\mathrm{With}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule} \\ $$$$\:\mathrm{L}=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2020x}^{\mathrm{2019}} −\mathrm{2021}}{\mathrm{3x}^{\mathrm{2}} } \\ $$$$\:=\frac{−\mathrm{2020}−\mathrm{2021}}{\mathrm{3}}=\:−\frac{\mathrm{4041}}{\mathrm{3}}=−\mathrm{1347} \\ $$

Commented by mathlove last updated on 21/Oct/22

with out L′Hopital rule sir

$${with}\:{out}\:{L}'{Hopital}\:{rule}\:{sir} \\ $$

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)

$$\:\mathrm{let}\:\mathrm{x}+\mathrm{1}=\mathrm{y} \\ $$$$\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2020}} −\mathrm{2021}\left(\mathrm{y}−\mathrm{1}\right)−\mathrm{2022}}{\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}} \\ $$$$=\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{2020}} −\mathrm{2021y}−\mathrm{1}}{\mathrm{y}^{\mathrm{3}} −\mathrm{3y}^{\mathrm{2}} +\mathrm{3y}} \\ $$$$=\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2020y}−\mathrm{2021y}−\mathrm{1}}{\mathrm{y}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{3y}+\mathrm{3}\right)} \\ $$$$=\:\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{4041y}}{\mathrm{y}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{3y}+\mathrm{3}\right)}=−\frac{\mathrm{4041}}{\mathrm{3}} \\ $$$$ \\ $$

Commented by mathlove last updated on 22/Oct/22

thank sir

$${thank}\:{sir} \\ $$

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

$$\:\:\:\:\:\:\:\underset{{x}\rightarrow\:−\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2020}} \:−\:\mathrm{1}\:−\:\mathrm{2021}{x}\:−\:\mathrm{2021}\:\:}{\left({x}\:+\:\mathrm{1}\right)\left(\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:=\underset{{x}\rightarrow\:−\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\:{x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{1019}} \:−\:{x}^{\mathrm{1018}} \:+\:{x}^{\mathrm{1017}} \:−\:{x}^{\mathrm{1016}} \:+\:.....\:+\:{x}\:−\:\mathrm{1}\right)\:−\:\mathrm{2021}\left({x}\:+\:\mathrm{1}\right)\:\:}{\left({x}\:+\:\mathrm{1}\right)\left(\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)}\: \\ $$$$\:\:\:\:\:\:=\underset{{x}\rightarrow\:−\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{1019}} \:−\:{x}^{\mathrm{1018}} \:+\:{x}^{\mathrm{1017}} \:−\:{x}^{\mathrm{1016}} \:+\:.....\:+\:{x}\:−\:\mathrm{1}\:−\:\mathrm{2021}}{\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}}\: \\ $$$$\:\:\:\:\:\:=\:\frac{−\mathrm{1}\:−\:\mathrm{1}\:−\:\mathrm{1}\:−\:\mathrm{1}\:−\:.....\:−\:\mathrm{1}\:−\:\mathrm{1}\:−\:\mathrm{2021}}{\:\mathrm{1}\:+\:\mathrm{1}\:+\:\mathrm{1}}\: \\ $$$$\:\:\:\:\:\:=\:\frac{−\:\mathrm{2020}\:−\:\mathrm{2021}}{\:\mathrm{3}}\: \\ $$$$\:\:\:\:\:\:=\:−\:\mathrm{1347} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mathlove last updated on 22/Oct/22

thanks sir

$${thanks}\:{sir} \\ $$

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