Question Number 12732 by Joel577 last updated on 30/Apr/17
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:−\:{x}}{\:\sqrt{{x}}\:+\:{x}} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17
$$=\mathrm{1} \\ $$
Commented by Joel577 last updated on 30/Apr/17
$${thank}\:{you} \\ $$
Answered by ridwan balatif last updated on 30/Apr/17
$$\frac{\sqrt{\mathrm{x}}−\mathrm{x}}{\:\sqrt{\mathrm{x}}+\mathrm{x}}×\frac{\sqrt{\mathrm{x}}+\mathrm{x}}{\:\sqrt{\mathrm{x}}+\mathrm{x}} \\ $$$$\frac{\mathrm{x}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2x}\sqrt{\mathrm{x}}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}\right)} \\ $$$$\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\mathrm{x}}{\:\sqrt{\mathrm{x}}+\mathrm{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{x}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{0}}+\mathrm{0}}=\mathrm{1} \\ $$
Commented by Joel577 last updated on 30/Apr/17
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by ajfour last updated on 30/Apr/17
$$\:{R}.{H}.{L}.\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\sqrt{{x}}\left(\mathrm{1}−\sqrt{{x}}\right)}{\:\sqrt{{x}}\left(\mathrm{1}+\sqrt{{x}}\right)}\: \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}} \\ $$$$=\mathrm{1}\: \\ $$
Commented by Joel577 last updated on 30/Apr/17
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by malwaan last updated on 30/Apr/17
$$\mathrm{put}\:{x}={y}^{\mathrm{2}} \\ $$$${x}\rightarrow\mathrm{0}\Rightarrow\mathrm{y}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }{\mathrm{y}+\mathrm{y}^{\mathrm{2}} }=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{y}\left(\mathrm{1}−\mathrm{y}\right)}{\mathrm{y}\left(\mathrm{1}+\mathrm{y}\right)} \\ $$$$=\underset{\mathrm{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{y}}{\mathrm{1}+\mathrm{y}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}+\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1} \\ $$
Commented by Joel577 last updated on 01/May/17
$${thank}\:{you}\:{very}\:{much} \\ $$