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Question Number 11444 by Joel576 last updated on 26/Mar/17

$$\mathrm{If}\underset{x\to 0}{\lim }\frac{\sqrt{px+q}-2}{x}=1$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}p+q?$$

Answered by ajfour last updated on 26/Mar/17

$$\mathrm{then}\underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{q}}{(1+\mathrm{px}/\mathrm{q})}^{1/2}-2}{\mathrm{x}}=1$$$$\Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{q}}(1+\frac{\mathrm{px}}{2\mathrm{q}})-2}{\mathrm{x}}=1$$$$=\underset{x\to 0}{\lim }\frac{(\sqrt{\mathrm{q}}-2)+\left(\frac{\mathrm{px}}{2\sqrt{\mathrm{q}}}\right)}{\mathrm{x}}=1$$$$\Rightarrow \sqrt{\mathrm{q}}=2\mathrm{or}\mathrm{q}=4$$$$\mathrm{then}$$$$\frac{\mathrm{p}}{2\sqrt{\mathrm{q}}}=1\mathrm{as}\sqrt{\mathrm{q}}=2,\mathrm{p}=4$$$$\mathrm{p}+\mathrm{q}=8$$

Commented by Joel576 last updated on 26/Mar/17

$$thankyouverymuch$$

Answered by ridwan balatif last updated on 26/Mar/17

$$\underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{px}+\mathrm{q}}-2}{\mathrm{x}}=1$$$$\mathrm{test}\mathrm{limit}:\frac{\sqrt{\mathrm{p}\times 0+\mathrm{q}}-2}{0}=1\to \mathrm{this}\mathrm{is}\mathrm{Impossible}$$$$\mathrm{so}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{test}\mathrm{limit}\mathrm{should}\mathrm{be}\frac{0}{0}$$$$\sqrt{\mathrm{p}\times 0+\mathrm{q}}-2=0$$$$\sqrt{\mathrm{q}}-2=0$$$$\mathrm{q}=4$$$$\underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{px}+4}-2}{\mathrm{x}}=1$$$$\underset{x\to 0}{\lim }\left(\frac{(\sqrt{\mathrm{px}+4}-2)}{\mathrm{x}}\times \left(\frac{\sqrt{\mathrm{px}+4}+2}{\sqrt{\mathrm{px}+4}+2}\right)\right)=1$$$$\underset{x\to 0}{\lim }\left(\frac{\mathrm{px}+4-4}{\mathrm{x}(\sqrt{\mathrm{px}+4}+2}\right)=1$$$$\underset{x\to 0}{\lim }\frac{\mathrm{px}}{\mathrm{x}(\sqrt{\mathrm{px}+4}+2)}=1$$$$\underset{x\to 0}{\lim }\frac{\mathrm{p}}{\sqrt{\mathrm{px}+4}+2}=1$$$$\frac{\mathrm{p}}{\sqrt{\mathrm{p}\times 0+4}+2}=1$$$$\frac{\mathrm{p}}{2+2}=1$$$$\mathrm{p}=4$$$$\mathrm{q}=4$$$$\therefore \mathrm{p}+\mathrm{q}=8$$

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