limit (1 Viewer)

C6H12O6-Glucose · Jul 30, 2009

can someone pls help me with this limit

lim x -> inf sqrt((x^2)+cx) - sqrt((x^2)+dx)

thanks

tommykins · Jul 30, 2009

multiply by $\frac{\sqrt{x^2+cx}+\sqrt{x^2+dx}}{\sqrt{x^2+cx}+\sqrt{x^2+dx}}$

jet · Jul 30, 2009

Aquawhite · Jul 30, 2009

jetblack2007 said:
$\lim_{x \to \infty} \left (\sqrt{x^2 + cx} - \sqrt{x^2 + dx} \right ) \\ = \lim_{x \to \infty} \left (\frac{\left (\sqrt{x^2 + cx} - \sqrt{x^2 + dx} \right ) \left (\sqrt{x^2 + cx} + \sqrt{x^2 + dx}\right )}{\sqrt{x^2 + cx} + \sqrt{x^2 + dx}} \right ) \\ = \lim_{x \to \infty} \left (\frac{x^2 + cx -x^2 - dx}{\sqrt{x^2 + cx} + \sqrt{x^2 + dx}} \right ) \\ = \lim_{x \to \infty} \left (\frac{(c - d)x}{\sqrt{x^2 + cx} + \sqrt{x^2 + dx}} \right ) \\ = \lim_{x \to \infty} \left (\frac{c - d}{\sqrt{1 + \frac{c}{x}}+ \sqrt{1 + \frac{d}{x}}} \right ) \\ \text{As }x \to \infty, \frac{1}{x} \to 0 \\ \therefore \lim_{x \to \infty} \left (\sqrt{x^2 + cx} - \sqrt{x^2 + dx} \right ) = \lim_{x \to \infty} \left (\frac{c - d}{\sqrt{1 + \frac{c}{x}}+ \sqrt{1 + \frac{d}{x}}} \right ) \\ = \frac{c - d}{2}$

Impressive.

limit (1 Viewer)

C6H12O6-Glucose

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tommykins

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jet

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