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Question-212428




Question Number 212428 by Spillover last updated on 13/Oct/24
Answered by Yassine84 last updated on 13/Oct/24
let f(x)=(1+x^2 )(1+x^3 )(1+x^4 ) then  lim_1  (((1+x^2 )(1+x^3 )(1+x^4 )−8)/(x−1))=f′(1)=36    because f′(x)=2x(1+x^3 )(1+x^4 )+3x^2 (1+x^2 )(1+x^4 )+4x^3 (1+x^2 )(1+x^3 )
$${let}\:{f}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:{then} \\ $$$${li}\underset{\mathrm{1}} {{m}}\:\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)−\mathrm{8}}{{x}−\mathrm{1}}={f}'\left(\mathrm{1}\right)=\mathrm{36} \\ $$$$ \\ $$$${because}\:{f}'\left({x}\right)=\mathrm{2}{x}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)+\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)+\mathrm{4}{x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right) \\ $$$$ \\ $$
Answered by golsendro last updated on 13/Oct/24
 = lim_(x→1)  (((x−1)(x^8 +x^7 +2x^6 +3x^5 +4x^4 +5x^3 +6x^2 +7x+7))/(x−1))     = 1+1+2+3+4+5+6+7+7   = 8+(7/2)(8)=8+28=36
$$\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{8}} +\mathrm{x}^{\mathrm{7}} +\mathrm{2x}^{\mathrm{6}} +\mathrm{3x}^{\mathrm{5}} +\mathrm{4x}^{\mathrm{4}} +\mathrm{5x}^{\mathrm{3}} +\mathrm{6x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{7}\right)}{\mathrm{x}−\mathrm{1}}\: \\ $$$$\:\:=\:\mathrm{1}+\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{7} \\ $$$$\:=\:\mathrm{8}+\frac{\mathrm{7}}{\mathrm{2}}\left(\mathrm{8}\right)=\mathrm{8}+\mathrm{28}=\mathrm{36} \\ $$
Answered by Nadirhashim last updated on 13/Oct/24
expanig and devid  im(x^8 +x^7 −2x^6 +3x^5 +4x^4 +5x^3   +4x^4 +5x^3 +6x^2 +7x+7   when x=1 the answer=36
$$\boldsymbol{{expanig}}\:\boldsymbol{{and}}\:\boldsymbol{{devid}} \\ $$$$\boldsymbol{{im}}\left(\boldsymbol{{x}}^{\mathrm{8}} +{x}^{\mathrm{7}} −\mathrm{2}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} \right. \\ $$$$+\mathrm{4}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{7} \\ $$$$\:{when}\:{x}=\mathrm{1}\:{the}\:{answer}=\mathrm{36} \\ $$

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