What is the limit of $(sqrt(x^2+x)-x)$ as x approaches infinity?

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Answer 1 · 2016-02-17T01:15:24

$lim_(xrarroo)(sqrt(x^2+x)-x)=1/2$

The initial form for the limit is indeterminate $oo-oo$

So, use the conjugate.

$(sqrt(x^2+x)-x) = (sqrt(x^2+x)-x)/1 *(sqrt(x^2+x)+x)/(sqrt(x^2+x)+x)$

$= (x^2+x-x^2)/(sqrt(x^2+x)+x)$

$= x/(sqrt(x^2+x)+x)$

$lim_(xrarroo) x/(sqrt(x^2+x)+x)$ has indeterminate form $oo/oo$ , but we can factor and reduce.

We know that $sqrt(x^2)=absx$ , so for positive $x$ (which is all we are concerned about for a limit as $x$ increases without bound) we have

$x/(sqrt(x^2+x)+x) = x/(sqrt(x^2)sqrt(1+1/x)+x)$ $" "$ (for all $x != 0$ )

$= x/(xsqrt(1+1/x)+x)$ $" "$ (for $x > 0$ )

$= x/(x(sqrt(1+1/x)+1))$

$= 1/(sqrt(1+1/x)+1)$

$lim_(xrarroo)1/(sqrt(1+1/x)+1) = 1/(sqrt1+1)=1/2$

What is the limit of (sqrt(x^2+x)-x) (sqrt(x^2+x)-x) as x approaches infinity?