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Question Number 141779 by iloveisrael last updated on 23/May/21
lim_(x→0) (x^2 /(x+2−2(√(1+x)))) =?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{x}}}\:=? \\ $$
Answered by Willson last updated on 23/May/21
lim_(x→0) (x^2 /(x+2−2(√(1+x)))) = lim_(x→0) ((2x)/(1−(1/( (√(1+x)))))) = lim_(x→0) ((2x(√(1+x)))/( (√(1+x))−1)) = lim_(x→0) ((2x(√(1+x))((√(1+x))+1))/(((√(1+x))−1)((√(1+x))+1))) = lim_(x→0) ((2x(√(1+x))((√(1+x))+1))/x) = lim_(x→0) 2 (√(1+x))((√(1+x))+1) = 4
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}}}{\:\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}}\left(\sqrt{\mathrm{1}+\mathrm{x}}+\mathrm{1}\right)}{\left(\sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{1}\right)\left(\sqrt{\mathrm{1}+\mathrm{x}}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}}\left(\sqrt{\mathrm{1}+\mathrm{x}}+\mathrm{1}\right)}{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}2}\:\sqrt{\mathrm{1}+\mathrm{x}}\left(\sqrt{\mathrm{1}+\mathrm{x}}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4} \\ $$
Answered by cherokeesay last updated on 23/May/21
= lim_(x→0) ((x^2 (x + 2 + 2(√(1 + x))))/((x + 2)^2 −4(1 + x))) = lim_(x→0) ((x^2 (x + 2 + 2(√(1 + x))))/(x^2 + 4 −4x −4 −4x)) = 2 + 2 = 4
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \left({x}\:+\:\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{1}\:+\:{x}}\right)}{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{4}\left(\mathrm{1}\:+\:{x}\right)}\:= \\ $$$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \left({x}\:+\:\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{1}\:+\:{x}}\right)}{{x}^{\mathrm{2}} \:+\:\mathrm{4}\:−\mathrm{4}{x}\:−\mathrm{4}\:−\mathrm{4}{x}}\:=\:\mathrm{2}\:+\:\mathrm{2}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\:\: \\ $$
Answered by iloveisrael last updated on 23/May/21
Answered by mathmax by abdo last updated on 23/May/21
letf(x)=(x^2 /(x+2−2(√(1+x)))) ⇒lim_(x→0) f(x) =lim_(x→0) ((x^2 ( x+2+2(√(1+x))))/((x+2)^2 −4(1+x))) =lim_(x→0) ((x^2 (x+2+2(√(1+x))))/(x^2 +4x+4−4−4x)) =lim_(x→0) x+2+2(√(1+x))=2+2=4
$$\mathrm{letf}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \left(\:\mathrm{x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}\right)}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+\mathrm{x}\right)}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4x}+\mathrm{4}−\mathrm{4}−\mathrm{4x}} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}=\mathrm{2}+\mathrm{2}=\mathrm{4} \\ $$

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