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Question Number 6405 by FilupSmith last updated on 26/Jun/16

How would you find the limit of:  lim_(x→∞)  x^3 −(x−1)^3

$$\mathrm{How}\:\mathrm{would}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$

Commented by FilupSmith last updated on 26/Jun/16

Do you have to expand and simplify?  Or is there a simpler method?

$$\mathrm{Do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{expand}\:\mathrm{and}\:\mathrm{simplify}? \\ $$$$\mathrm{Or}\:\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{simpler}\:\mathrm{method}? \\ $$

Commented by Rasheed Soomro last updated on 27/Jun/16

lim_(x→∞)  x^3 −(x−1)^3 =?    x^3 −(x−1)^3        =x^3 −x^3 +3x^2 −3x+1      =3x^2 −3x+1       =3x^2 −6x+3+3x−2       =3(x^2 −2x+1)+3x−2       =3(x−1)^2 +3x−2    lim_(x→∞)  x^3 −(x−1)^3 =lim_(x→∞) (3(x−1)^2 +3x−2)     =3(∞−1)^2 +3(∞)−2=∞

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} =? \\ $$$$ \\ $$$${x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:={x}^{\mathrm{3}} −{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\:\:\:=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\:\:\:\:=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{3}+\mathrm{3}{x}−\mathrm{2} \\ $$$$\:\:\:\:\:=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{3}{x}−\mathrm{2} \\ $$$$\:\:\:\:\:=\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}\right)\:\:\: \\ $$$$=\mathrm{3}\left(\infty−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left(\infty\right)−\mathrm{2}=\infty \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 27/Jun/16

lim_(x→∞)  x^3 −(x−1)^3 =?      x^3 −(x−1)^3              =(x−(x−1))(x^2 +x(x−1)+(x−1)^2 )             =x^2 +x(x−1)+(x−1)^2     lim_(x→∞)  x^3 −(x−1)^3 =lim_(x→∞)  ( x^2 +x(x−1)+(x−1)^2  )  ∞^2 +∞(∞−1)+(∞−1)^2   =∞+∞+∞=∞

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} =? \\ $$$$ \\ $$$$\:\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left({x}−\left({x}−\mathrm{1}\right)\right)\left({x}^{\mathrm{2}} +{x}\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} +{x}\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{3}} −\left({x}−\mathrm{1}\right)^{\mathrm{3}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\:{x}^{\mathrm{2}} +{x}\left({x}−\mathrm{1}\right)+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\right) \\ $$$$\infty^{\mathrm{2}} +\infty\left(\infty−\mathrm{1}\right)+\left(\infty−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\infty+\infty+\infty=\infty \\ $$

Commented by Rasheed Soomro last updated on 27/Jun/16

Don′t know simpler method.

$${Don}'{t}\:{know}\:{simpler}\:{method}. \\ $$

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