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Question Number 146505 by mathdanisur last updated on 13/Jul/21

lim_(x→7) (((√(x - a)) - 4)/(x - 7)) = b find a∙b=?

$$\underset{{x}\rightarrow\mathrm{7}} {{lim}}\:\frac{\sqrt{{x}\:-\:{a}}\:-\:\mathrm{4}}{{x}\:-\:\mathrm{7}}\:=\:{b}\:\:{find}\:\:{a}\centerdot{b}=? \\ $$

Answered by ajfour last updated on 13/Jul/21

i just dunno but letsee whtits bout.. if limit has to exist, i.e. b finite then d^r →0 ⇒ n^r →0 so x−a=16 but x→7 ⇒ a→ −9 but then b= ((∂n^r /∂x)/(∂d^r /∂x))=((1/(2(√(x−a))))/1) but gain x→7, so ⇒ b=(1/(2(√(7−(−9)))))=(1/8) now if you ask ab forget (∙), u have −(9/8)

$${i}\:{just}\:{dunno}\:\:{but}\:{letsee}\:{whtits} \\ $$$${bout}.. \\ $$$${if}\:{limit}\:{has}\:{to}\:{exist},\:{i}.{e}. \\ $$$${b}\:{finite}\:{then}\:{d}^{{r}} \rightarrow\mathrm{0}\:\:\Rightarrow\:{n}^{{r}} \rightarrow\mathrm{0} \\ $$$${so}\:\:\:\:{x}−{a}=\mathrm{16} \\ $$$${but}\:{x}\rightarrow\mathrm{7}\:\:\Rightarrow\:\:\:{a}\rightarrow\:−\mathrm{9} \\ $$$${but}\:{then} \\ $$$$\:\:{b}=\:\frac{\frac{\partial{n}^{{r}} }{\partial{x}}}{\frac{\partial{d}^{{r}} }{\partial{x}}}=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}−{a}}}}{\mathrm{1}} \\ $$$${but}\:{gain}\:{x}\rightarrow\mathrm{7},\:{so} \\ $$$$\Rightarrow\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}−\left(−\mathrm{9}\right)}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${now}\:{if}\:{you}\:{ask} \\ $$$$\:\:{ab}\:\:\:{forget}\:\left(\centerdot\right),\:\:{u}\:{have}\:−\frac{\mathrm{9}}{\mathrm{8}} \\ $$

Commented by mathdanisur last updated on 13/Jul/21

thanks Ser

$${thanks}\:{Ser} \\ $$

Answered by mathmax by abdo last updated on 13/Jul/21

f(x)=(((√(x−a))−4)/(x−7)) changement x−7=t give f(x)=(((√(t+7−a))−4)/t)=g(t) (t→0) ⇒ g(t)=(((√(7−a))(√(1+(t/(7−a))))−4)/t)∼(((√(7−a))×(1+(t/(2(7−a))))−4)/t) (((√(7−a))−4)/t) +(1/(2(√(7−a))))=b ⇒(√(7−a))−4=0 ⇒7−a=16 ⇒a=−9 ⇒ b=(1/(2(√(7+9))))=(1/8) ⇒ab=−(9/8)

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}−\mathrm{a}}−\mathrm{4}}{\mathrm{x}−\mathrm{7}}\:\:\mathrm{changement}\:\mathrm{x}−\mathrm{7}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{t}+\mathrm{7}−\mathrm{a}}−\mathrm{4}}{\mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right)\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{\sqrt{\mathrm{7}−\mathrm{a}}\sqrt{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{7}−\mathrm{a}}}−\mathrm{4}}{\mathrm{t}}\sim\frac{\sqrt{\mathrm{7}−\mathrm{a}}×\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}\left(\mathrm{7}−\mathrm{a}\right)}\right)−\mathrm{4}}{\mathrm{t}} \\ $$$$\frac{\sqrt{\mathrm{7}−\mathrm{a}}−\mathrm{4}}{\mathrm{t}}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}−\mathrm{a}}}=\mathrm{b}\:\:\Rightarrow\sqrt{\mathrm{7}−\mathrm{a}}−\mathrm{4}=\mathrm{0}\:\Rightarrow\mathrm{7}−\mathrm{a}=\mathrm{16}\:\Rightarrow\mathrm{a}=−\mathrm{9}\:\Rightarrow \\ $$$$\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{7}+\mathrm{9}}}=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\mathrm{ab}=−\frac{\mathrm{9}}{\mathrm{8}} \\ $$

Commented by mathdanisur last updated on 13/Jul/21

thank you Ser

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

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