Limit Calculation

**WeeG** · May 11th 2010, 02:03 AM

Hi guys, I need a little help with this limit here:

lim x->infinity of: sqrt(x^2 + x) - sqrt(x^2 - 2x)

cheers !

**HallsofIvy** · May 11th 2010, 02:22 AM

Originally Posted by WeeG

Hi guys, I need a little help with this limit here:

lim x->infinity of: sqrt(x^2 + x) - sqrt(x^2 - 2x)

cheers !

"Rationalize the numerator"- that is, multiply both numerator and denominator by $\sqrt{x^2+ x}+ \sqrt{x^2- 2x}$ to get $\frac{(\sqrt{x^2+ x}- \sqrt{x^2- 2x})(\sqrt{x^2+ x}+ \sqrt{x^2- 2x})}{\sqrt{x^2+ x}+ \sqrt{x^2- 2x}}$ $= \frac{x^2+ x- x^2+ 2x}{\sqrt{x^2+x}+ \sqrt{x^2- 2x}}= \frac{3x}{\sqrt{x^2+ x}+ \sqrt{x^2- 2x}}$

**WeeG** · May 11th 2010, 06:01 AM

done it, cheers !

**simplependulum** · May 11th 2010, 06:24 AM

Or we can solve it by mean value theorem ,

Let $f(x) = \sqrt{t}$
we have

$\frac{ \sqrt{x^2 + x } - \sqrt{x^2 - 2x}}{x^2 + x - x^2 + 2x} = \frac{1}{2\sqrt{c}}$ where $x^2 - 2x < c < x^2 + x$

Take reciporcal and square root ,

$\frac{1}{\sqrt{x^2 - 2x}} > \frac{1}{\sqrt{c}} > \frac{1}{\sqrt{x^2 +x }}$

$\frac{1}{\sqrt{x^2 - 2x}} > 2 \frac{ \sqrt{x^2 + x } - \sqrt{x^2 - 2x}}{3x}> \frac{1}{\sqrt{x^2 +x }}$

$\frac{3}{2} \frac{x}{\sqrt{x^2 - 2x}} > \sqrt{x^2 + x } - \sqrt{x^2 - 2x} > \frac{3}{2} \frac{x}{\sqrt{x^2 +x }}$

$\lim_{x\to\infty} \frac{3}{2} \frac{x}{\sqrt{x^2 - 2x}}> \lim_{x\to\infty} \sqrt{x^2 + x } - \sqrt{x^2 - 2x} > \lim_{x\to\infty} \frac{3}{2} \frac{x}{\sqrt{x^2 +x }}$

$\frac{3}{2} > \lim_{x\to\infty} \sqrt{x^2 + x } - \sqrt{x^2 - 2x} >\frac{3}{2}$

$L = \frac{3}{2}$

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