If-lim-x-0-px-q-2-x-1-What-is-the-value-of-p-q-

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

If \lim_{x \to 0} \frac{\sqrt{p x + q} - 2}{x} = 1

What is the value of p + q ?

Answered by ajfour last updated on 26/Mar/17

then \lim_{x \to 0} \frac{\sqrt{q} {(1 + px / q)}^{1 / 2} - 2}{x} = 1

\Rightarrow \lim_{x \to 0} \frac{\sqrt{q} (1 + \frac{px}{2 q}) - 2}{x} = 1

= \lim_{x \to 0} \frac{(\sqrt{q} - 2) + (\frac{px}{2 \sqrt{q}})}{x} = 1

\Rightarrow \sqrt{q} = 2 or q = 4

then

\frac{p}{2 \sqrt{q}} = 1 as \sqrt{q} = 2, p = 4

p + q = 8

Commented by Joel576 last updated on 26/Mar/17

t h a n k y o u v e r y m u c h

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

\lim_{x \to 0} \frac{\sqrt{px + q} - 2}{x} = 1

test limit : \frac{\sqrt{p \times 0 + q} - 2}{0} = 1 \to this is Impossible

so form of the test limit should be \frac{0}{0}

\sqrt{p \times 0 + q} - 2 = 0

\sqrt{q} - 2 = 0

q = 4

\lim_{x \to 0} \frac{\sqrt{px + 4} - 2}{x} = 1

\lim_{x \to 0} (\frac{(\sqrt{px + 4} - 2)}{x} \times (\frac{\sqrt{px + 4} + 2}{\sqrt{px + 4} + 2})) = 1

\lim_{x \to 0} (\frac{px + 4 - 4}{x (\sqrt{px + 4} + 2}) = 1

\lim_{x \to 0} \frac{px}{x (\sqrt{px + 4} + 2)} = 1

\lim_{x \to 0} \frac{p}{\sqrt{px + 4} + 2} = 1

\frac{p}{\sqrt{p \times 0 + 4} + 2} = 1

\frac{p}{2 + 2} = 1

p = 4

q = 4

∴ p + q = 8