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Question Number 104533 by 175mohamed last updated on 22/Jul/20

find : lim_(x→1) (((x+2)^2 +(x+1)^3 −17)/((√(x+3))−((x+7))^(1/3) ))

$${find}\:: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{17}}{\sqrt{{x}+\mathrm{3}}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{7}}} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Jul/20

lim_(x→1) ((2(x+2)+3(x+1)^2 )/((1/2).(1/(√(x+3)))−(1/3)(x+7)^(−(2/3)) ))=((18)/((1/4)−(1/(12))))=108

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}+\mathrm{2}\right)+\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\sqrt{{x}+\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{7}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{18}}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{12}}}=\mathrm{108} \\ $$

Answered by OlafThorendsen last updated on 22/Jul/20

x = u+1 lim_(u→0) (((u+3)^2 +(u+2)^3 −17)/((√(u+4))−((u+8))^(1/3) )) lim_(u→0) ((9(1+(u/3))^2 +8(1+(u/2))^3 −17)/(2(√(1+(u/4)))−2((1+(u/8)))^(1/3) )) lim_(u→0) ((9(1+((2u)/3))+8(1+((3u)/2))−17)/(2(1+(u/8))−2(1+(u/(24))))) lim_(u→0) ((9(((2u)/3))+8(((3u)/2)))/(2((u/8))−2((u/(24))))) lim_(u→0) ((6u+12u)/((u/4)−(u/(12)))) lim_(u→0) ((18u)/(u/6)) = 18×6 = 108

$${x}\:=\:{u}+\mathrm{1} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({u}+\mathrm{3}\right)^{\mathrm{2}} +\left({u}+\mathrm{2}\right)^{\mathrm{3}} −\mathrm{17}}{\sqrt{{u}+\mathrm{4}}−\sqrt[{\mathrm{3}}]{{u}+\mathrm{8}}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{9}\left(\mathrm{1}+\frac{{u}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{8}\left(\mathrm{1}+\frac{{u}}{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{17}}{\mathrm{2}\sqrt{\mathrm{1}+\frac{{u}}{\mathrm{4}}}−\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{{u}}{\mathrm{8}}}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{9}\left(\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{3}}\right)+\mathrm{8}\left(\mathrm{1}+\frac{\mathrm{3}{u}}{\mathrm{2}}\right)−\mathrm{17}}{\mathrm{2}\left(\mathrm{1}+\frac{{u}}{\mathrm{8}}\right)−\mathrm{2}\left(\mathrm{1}+\frac{{u}}{\mathrm{24}}\right)} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{9}\left(\frac{\mathrm{2}{u}}{\mathrm{3}}\right)+\mathrm{8}\left(\frac{\mathrm{3}{u}}{\mathrm{2}}\right)}{\mathrm{2}\left(\frac{{u}}{\mathrm{8}}\right)−\mathrm{2}\left(\frac{{u}}{\mathrm{24}}\right)} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{6}{u}+\mathrm{12}{u}}{\frac{{u}}{\mathrm{4}}−\frac{{u}}{\mathrm{12}}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{18}{u}}{\frac{{u}}{\mathrm{6}}}\:=\:\mathrm{18}×\mathrm{6}\:=\:\mathrm{108} \\ $$$$ \\ $$

Commented by bobhans last updated on 22/Jul/20

(u/4)−(u/(12)) = ((3u)/(12))−(u/(12)) = ((2u)/(12)) = (u/6)

$$\frac{{u}}{\mathrm{4}}−\frac{{u}}{\mathrm{12}}\:=\:\frac{\mathrm{3}{u}}{\mathrm{12}}−\frac{{u}}{\mathrm{12}}\:=\:\frac{\mathrm{2}{u}}{\mathrm{12}}\:=\:\frac{{u}}{\mathrm{6}} \\ $$

Answered by bramlex last updated on 22/Jul/20

L′Hopital rule lim_(x→1) ((2(x+2)+3(x+1)^2 )/((1/(2(√(x+3))))−(1/(3 (((x+7)^2 ))^(1/3) )))) = ((6+12)/((1/4)−(1/(3.4)))) = ((18)/(((1/6)))) = 108 □

$${L}'{Hopital}\:{rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}+\mathrm{2}\right)+\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{7}\right)^{\mathrm{2}} }}}\:= \\ $$$$\frac{\mathrm{6}+\mathrm{12}}{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}}\:=\:\frac{\mathrm{18}}{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}\:=\:\mathrm{108}\:\square \\ $$

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