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Question Number 121576 by oustmuchiya@gmail.com last updated on 09/Nov/20

Answered by TANMAY PANACEA last updated on 09/Nov/20

f(x)=((((√(x^2 +15)) −5)((√(x^2 +15)) +5))/((x−2)((√(x^2 +15)) +5))) =((x^2 +15−25)/((x−2)((√(x^2 +15)) +5)))=((x^2 −10)/((x−2)((√(x^2 +15)) +5))) at x=2 the function[discontinous lim_(x→2) f(x)=∞

$${f}\left({x}\right)=\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:−\mathrm{5}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{15}−\mathrm{25}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)}=\frac{{x}^{\mathrm{2}} −\mathrm{10}}{\left({x}−\mathrm{2}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{15}}\:+\mathrm{5}\right)} \\ $$$${at}\:{x}=\mathrm{2}\:{the}\:{function}\left[{discontinous}\right. \\ $$$${li}\underset{{x}\rightarrow\mathrm{2}} {{m}f}\left({x}\right)=\infty \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 09/Nov/20

f(x)=((x(√(1+((15)/x^2 ))) −5)/(x−2))=(((√(1+((15)/x^2 ))) −(5/x))/(1−(2/x))) so lim_(x→∞) f(x) =(((√(1+0)) −0)/(1−0))=1

$${f}\left({x}\right)=\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{15}}{{x}^{\mathrm{2}} }}\:−\mathrm{5}}{{x}−\mathrm{2}}=\frac{\sqrt{\mathrm{1}+\frac{\mathrm{15}}{{x}^{\mathrm{2}} }}\:−\frac{\mathrm{5}}{{x}}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}} \\ $$$${so}\:{li}\underset{{x}\rightarrow\infty} {{m}}\:{f}\left({x}\right) \\ $$$$=\frac{\sqrt{\mathrm{1}+\mathrm{0}}\:−\mathrm{0}}{\mathrm{1}−\mathrm{0}}=\mathrm{1} \\ $$

Commented by oustmuchiya@gmail.com last updated on 12/Nov/20

many thanks indeed

$$\mathrm{many}\:\mathrm{thanks}\:\mathrm{indeed} \\ $$

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