Question Number 110988 by Rio Michael last updated on 01/Sep/20
$$\:\mathrm{Evaluate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{L}'\mathrm{hopital}'\mathrm{s}\:\mathrm{rule} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$
Answered by Rasheed.Sindhi last updated on 01/Sep/20
$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$$$\:\:\:\frac{\sqrt{{x}}−\mathrm{2}}{\:\left(\sqrt{{x}}\right)^{\mathrm{2}} −\left(\mathrm{2}\right)^{\mathrm{2}} }=\frac{\sqrt{{x}}−\mathrm{2}}{\:\left(\sqrt{{x}}−\mathrm{2}\right)\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}=\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{1}}{\:\left(\sqrt{{x}}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}+\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Rio Michael last updated on 01/Sep/20
$$\mathrm{correct}\:\mathrm{sir}. \\ $$
Answered by malwan last updated on 02/Sep/20
$$\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}×\frac{\sqrt{{x}}+\mathrm{2}}{\:\sqrt{{x}}+\mathrm{2}} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{{x}−\mathrm{4}}{\left({x}−\mathrm{4}\right)\left(\sqrt{{x}}+\mathrm{2}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{1}}{\:\sqrt{{x}}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by bemath last updated on 01/Sep/20
$$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{b}\:\Rightarrow\mathrm{x}\:=\:\mathrm{b}^{\mathrm{2}} \\ $$$$\underset{\mathrm{b}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{b}−\mathrm{2}}{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}\:=\:\underset{\mathrm{b}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{b}+\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$