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Question Number 9654 by ridwan balatif last updated on 22/Dec/16

lim_(x→4) ((ax+b−(√x))/(x−4))=(3/4), a+b=...?

$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{ax}+\mathrm{b}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}},\:\:\mathrm{a}+\mathrm{b}=...? \\ $$

Answered by prakash jain last updated on 22/Dec/16

lim_(x→4) ((ax+b−(√x))/(x−4))=(3/4), a+b=...? Since limit exists lim_(x→4) ax+b−(√x)=0 4a+b−2=0 b=2−4a ((ax+2−4a−(√x))/(x−4))=((a(x−4)−((√x)−2))/(x−4)) =((a(x−4)−(((x−4))/(((√x)+2))))/(x−4))=a−(1/((√x)+2)) a−(1/4)=(3/4)⇒a=1 b=2−4a=−2 a+b=−1 check lim_(x→4) ((x−2−(√x))/(x−4))=lim_(x→4) (((x−4)−((√x)−2))/((x−4))) =lim_(x→4) (((x−4)(1−(1/((√x)+2))))/((x−4)))=(3/4)

$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{ax}+\mathrm{b}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}},\:\:\mathrm{a}+\mathrm{b}=...? \\ $$$$\mathrm{Since}\:\mathrm{limit}\:\mathrm{exists} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\mathrm{ax}+\mathrm{b}−\sqrt{\mathrm{x}}=\mathrm{0} \\ $$$$\mathrm{4a}+\mathrm{b}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{b}=\mathrm{2}−\mathrm{4a} \\ $$$$\frac{\mathrm{ax}+\mathrm{2}−\mathrm{4a}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\frac{\mathrm{a}\left(\mathrm{x}−\mathrm{4}\right)−\left(\sqrt{{x}}−\mathrm{2}\right)}{\mathrm{x}−\mathrm{4}} \\ $$$$=\frac{\mathrm{a}\left(\mathrm{x}−\mathrm{4}\right)−\frac{\left(\mathrm{x}−\mathrm{4}\right)}{\left(\sqrt{\mathrm{x}}+\mathrm{2}\right)}}{\mathrm{x}−\mathrm{4}}=\mathrm{a}−\frac{\mathrm{1}}{\sqrt{\mathrm{x}}+\mathrm{2}} \\ $$$$\mathrm{a}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow{a}=\mathrm{1} \\ $$$${b}=\mathrm{2}−\mathrm{4}{a}=−\mathrm{2} \\ $$$${a}+{b}=−\mathrm{1} \\ $$$${check} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{x}−\mathrm{2}−\sqrt{\mathrm{x}}}{\mathrm{x}−\mathrm{4}}=\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\left({x}−\mathrm{4}\right)−\left(\sqrt{{x}}−\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{4}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{{x}}+\mathrm{2}}\right)}{\left({x}−\mathrm{4}\right)}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$

Commented by ridwan balatif last updated on 23/Dec/16

thank you sir for your help

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help} \\ $$

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