Question Number 200060 by sonukgindia last updated on 13/Nov/23
Answered by cortano12 last updated on 13/Nov/23
$$\:\mathrm{L}=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\mathrm{2}{ab}−{a}\sqrt{{ab}}−{ab}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ab}−{a}\sqrt{{ab}}}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)} \\ $$$$\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\sqrt{{ab}}\:\left(\sqrt{{ab}}−{a}\right)}{\left({a}+\sqrt{{ab}}\right)\left({b}−{a}\right)}\: \\ $$$$\:=\:\frac{{a}}{\mathrm{2}{a}}\:.\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\sqrt{{a}}\:\left(\sqrt{{b}}−\sqrt{{a}}\:\right)}{\left(\sqrt{{b}}+\sqrt{{a}}\right)\left(\sqrt{{b}}−\sqrt{{a}}\:\right)} \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:.\:\frac{\sqrt{{a}}}{\mathrm{2}\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by Frix last updated on 13/Nov/23
$$\mathrm{Let}\:{b}={at}^{\mathrm{2}} \wedge{a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge{t}>\mathrm{0} \\ $$$${b}\rightarrow{a}\:\Rightarrow\:{t}\rightarrow\mathrm{1} \\ $$$$\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{ab}}{{a}+\sqrt{{ab}}}−\sqrt{{ab}}}{{b}−{a}}\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{t}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{If}\:{a}<\mathrm{0}\wedge{b}<\mathrm{0}\wedge{t}>\mathrm{0} \\ $$$$\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{ab}}{{a}+\sqrt{{ab}}}−\sqrt{{ab}}}{{b}−{a}}\:=\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−{t}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:=−\infty \\ $$