Find the limit of the following term:
$$\lim_{x\to\infty} \sqrt{2x-3}-\sqrt{ax+b},$$
while $a\in\mathbb{R}^+,b\in\mathbb{R}$.
Please, I need help!!!
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1Tell us what you already know to help us better help you. Do you know what a limit is? Do you know what $\lim_{x\to\infty}\sqrt{2x-3}$ is? – JRN Apr 28 '15 at 13:12
2 Answers
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Hint: Multiply and divide by the conjugate.
Travis Willse
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Vincenzo Oliva
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...and then see for what values of $a$ this converges. What about $b$, does its value matter? – Simon S Apr 28 '15 at 12:32
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My comment was intended as an additional hint for the OPer (and not as a correction for you or anything else negative). – Simon S Apr 28 '15 at 14:37
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*OPer; @SimonS Oh, I see. I thought you were suggesting an improvement, so I wanted to clarify my intention. No problem anyway. – Vincenzo Oliva Apr 28 '15 at 14:44
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$$\sqrt{2x-3}-\sqrt{ax+b} = \dfrac{(\sqrt{2x-3} - \sqrt{ax+b})(\sqrt{2x+3} + \sqrt{ax+b})}{\sqrt{2x-3} + \sqrt{ax+b}} =\frac{(2x-3)-(ax+b)}{\sqrt{2x-3} + \sqrt{ax+b}}$$
$$= \frac{(2-a)x-(3+b)}{\sqrt{2x-3}+\sqrt{ax+b}}$$
Can you take the limit now? The degree of the numerator exceeds that of the denominator: try dividing numerator and denominator by $\sqrt x$. Then use the fact that given $a>0$, $2-a>0$ provided $0\leq a \lt 2$, $2-a = 0$ when $a=2$, and $2-a \lt 0$ when $a>2$.
Jordan Glen
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