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Question Number 12732 by Joel577 last updated on 30/Apr/17

lim_(x→0) (((√x) − x)/((√x) + x))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:−\:{x}}{\sqrt{{x}}\:+\:{x}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17

=1

$$=\mathrm{1} \\ $$

Commented by Joel577 last updated on 30/Apr/17

thank you

$${thank}\:{you} \\ $$

Answered by ridwan balatif last updated on 30/Apr/17

(((√x)−x)/((√x)+x))×(((√x)+x)/((√x)+x)) ((x−x^2 )/(x+2x(√x)+x^2 )) ((x(1−x))/(x(1+2(√x)+x))) ((1−x)/(1+2(√x)+x)) lim_(x→0) (((√x)−x)/((√x)+x))=lim_(x→0) ((1−x)/(1+2(√x)+x))=((1−0)/(1+2(√0)+0))=1

$$\frac{\sqrt{\mathrm{x}}−\mathrm{x}}{\sqrt{\mathrm{x}}+\mathrm{x}}×\frac{\sqrt{\mathrm{x}}+\mathrm{x}}{\sqrt{\mathrm{x}}+\mathrm{x}} \\ $$$$\frac{\mathrm{x}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2x}\sqrt{\mathrm{x}}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}\right)} \\ $$$$\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\mathrm{x}}{\sqrt{\mathrm{x}}+\mathrm{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{0}}+\mathrm{0}}=\mathrm{1} \\ $$

Commented by Joel577 last updated on 30/Apr/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by ajfour last updated on 30/Apr/17

R.H.L. =lim_(x→0) (((√x)(1−(√x)))/((√x)(1+(√x)))) =lim_(x→0) ((1−(√x))/(1+(√x))) =1

$$\:{R}.{H}.{L}.\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\sqrt{{x}}\left(\mathrm{1}−\sqrt{{x}}\right)}{\sqrt{{x}}\left(\mathrm{1}+\sqrt{{x}}\right)}\: \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}} \\ $$$$=\mathrm{1}\: \\ $$

Commented by Joel577 last updated on 30/Apr/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by malwaan last updated on 30/Apr/17

put x=y^2 x→0⇒y→0 lim_(y→0) ((y−y^2 )/(y+y^2 ))=lim_(y→0) ((y(1−y))/(y(1+y))) =lim_(y→0) ((1−y)/(1+y))=((1−0)/(1+0))=(1/1)=1

$$\mathrm{put}\:{x}={y}^{\mathrm{2}} \\ $$$${x}\rightarrow\mathrm{0}\Rightarrow\mathrm{y}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }{\mathrm{y}+\mathrm{y}^{\mathrm{2}} }=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{y}\left(\mathrm{1}−\mathrm{y}\right)}{\mathrm{y}\left(\mathrm{1}+\mathrm{y}\right)} \\ $$$$=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{y}}{\mathrm{1}+\mathrm{y}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}+\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1} \\ $$

Commented by Joel577 last updated on 01/May/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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