Question Number 84330 by M±th+et£s last updated on 11/Mar/20
$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)−\mathrm{64}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$
Commented by M±th+et£s last updated on 11/Mar/20
$${typo}\:\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$
Commented by niroj last updated on 12/Mar/20
$$\:\underset{\mathrm{x}\rightarrow\mathrm{1}} {\overset{\mathrm{lim}} {\:}}\:\frac{\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{64}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{1}} {\overset{\mathrm{lim}} {\:}}\:\:\frac{\mathrm{3}\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{6x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\left(\mathrm{2}.\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} .\left(\mathrm{6}.\mathrm{1}+\mathrm{1}\right)}{\mathrm{1}}=\mathrm{16}.\mathrm{7} \\ $$$$\:=\:\mathrm{112}.\:\:\mathrm{now}\:\mathrm{correlation}. \\ $$$$ \\ $$
Commented by jagoll last updated on 11/Mar/20
$$\mathrm{wrong}\:\mathrm{sir} \\ $$
Answered by jagoll last updated on 12/Mar/20
![(2x^3 +x+1)^3 −4^3 = (2x^3 +x+1−4)[(2x^3 +x+1)^2 +4(2x^3 +x+1)+16 ] lim_(x→1) [ (2x^3 +x+1)^2 +4(2x^3 +3+1)+16 ] × lim_(x→1) ((2x^3 +x−3)/(x^3 −1)) = 48 × lim_(x→1) (((x−1)(2x^2 +2x+3))/((x−1)(x^3 +x+1))) = 48 × (7/3) = 16×7 = 112](https://www.tinkutara.com/question/Q84353.png)
$$\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} \:=\: \\ $$$$\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}−\mathrm{4}\right)\left[\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)+\mathrm{16}\:\right] \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left[\:\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2x}^{\mathrm{3}} +\mathrm{3}+\mathrm{1}\right)+\mathrm{16}\:\right]\:× \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{3}} +\mathrm{x}−\mathrm{3}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}\:=\: \\ $$$$\mathrm{48}\:×\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}\right)}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}\right)}\:=\: \\ $$$$\mathrm{48}\:×\:\frac{\mathrm{7}}{\mathrm{3}}\:=\:\mathrm{16}×\mathrm{7}\:=\:\mathrm{112} \\ $$
Commented by naka3546 last updated on 12/Mar/20
$${why}\:\:{x}\:\rightarrow\:\:\mathrm{0}\:\:? \\ $$$${second}\:\:{line}\:\:{from}\:\:{bottom}\:\:{line}\:. \\ $$
Commented by jagoll last updated on 12/Mar/20
$$\mathrm{x}\rightarrow\mathrm{1} \\ $$