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Question Number 140071 by EDWIN88 last updated on 04/May/21
For what value of k is the following  continous function ?  f(x)= { (((((√(7x+2))−(√(6x+4)))/(x−2)) ; if x≥−(2/7) & x≠2)),((        k                       ; if x=2)) :}
$$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{is}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{continous}\:\mathrm{function}\:? \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\frac{\sqrt{\mathrm{7x}+\mathrm{2}}−\sqrt{\mathrm{6x}+\mathrm{4}}}{\mathrm{x}−\mathrm{2}}\:;\:\mathrm{if}\:\mathrm{x}\geqslant−\frac{\mathrm{2}}{\mathrm{7}}\:\&\:\mathrm{x}\neq\mathrm{2}}\\{\:\:\:\:\:\:\:\:\mathrm{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:;\:\mathrm{if}\:\mathrm{x}=\mathrm{2}}\end{cases} \\ $$
Answered by bobhans last updated on 04/May/21
 lim_(x→2)  f(x)=lim_(x→2)  (((√(7x+2))−(√(6x+4)))/(x−2))  = lim_(x→2)  (((x−2))/((x−2)((√(7x+2))+(√(6x+4)))))  = lim_(x→2)  (1/( (√(7x+2))+(√(6x+4)))) = (1/8)=k
$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{7x}+\mathrm{2}}−\sqrt{\mathrm{6x}+\mathrm{4}}}{\mathrm{x}−\mathrm{2}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\mathrm{2}\right)}{\left(\mathrm{x}−\mathrm{2}\right)\left(\sqrt{\mathrm{7x}+\mathrm{2}}+\sqrt{\mathrm{6x}+\mathrm{4}}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7x}+\mathrm{2}}+\sqrt{\mathrm{6x}+\mathrm{4}}}\:=\:\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{k} \\ $$

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