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Question Number 93676 by john santu last updated on 14/May/20

solve without L′Hopital lim_(x→0) ((5(√(x+1))−2(√(x+4))−1)/x) ?

$$\mathrm{solve}\:\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{x}}\:? \\ $$

Commented by mathmax by abdo last updated on 14/May/20

let f(x) =((5(√(x+1))−2(√(x+4))−1)/x) we have (√(1+x))∼1+(x/2) (√(x+4))=2(√(1+(x/4)))∼2(1+(x/8)) ⇒f(x) ∼((5(1+(x/2))−2(2+(x/4))−1)/x) =((5+((5x)/2)−4−(x/2)−1)/x) =((2x)/x) =2 ⇒lim_(x→0) f(x) =2

$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{5}\sqrt{{x}+\mathrm{1}}−\mathrm{2}\sqrt{{x}+\mathrm{4}}−\mathrm{1}}{{x}}\:\:{we}\:{have}\:\sqrt{\mathrm{1}+{x}}\sim\mathrm{1}+\frac{{x}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{4}}=\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}}{\mathrm{4}}}\sim\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{8}}\right)\:\Rightarrow{f}\left({x}\right)\:\sim\frac{\mathrm{5}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}\left(\mathrm{2}+\frac{{x}}{\mathrm{4}}\right)−\mathrm{1}}{{x}} \\ $$$$=\frac{\mathrm{5}+\frac{\mathrm{5}{x}}{\mathrm{2}}−\mathrm{4}−\frac{{x}}{\mathrm{2}}−\mathrm{1}}{{x}}\:=\frac{\mathrm{2}{x}}{{x}}\:=\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)\:=\mathrm{2} \\ $$

Answered by i jagooll last updated on 14/May/20

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