How do you find the limit of $(sqrt(x^2-1) ) / (sqrt(4x^2+x) )$ as x approaches infinity?

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Answer 1 · 2016-05-05T18:54:27

The limit is $1/2$ . For "how", please see the explanation section below.

In investigating a limit as $x$ increases without bounds, we are not concerned about what happens at $x=0$ .

We note that for all $x != 0$ ,

$sqrt(x^2-1)/sqrt(4x^2+x) = sqrt(x^2(1-1/x^2))/sqrt(x^2(4+1/x))$

$= (sqrt(x^2)sqrt(1-1/x^2))/(sqrt(x^2)sqrt(4-1/x))$

Note also that $sqrt(x^2) = absx$ , so for positive $x$ , we have $sqrt(x^2) = x$ .

(For negative $x$ , we get $sqrt(x^2) = -x$ )

We continue:

$sqrt(x^2-1)/sqrt(4x^2+x) = (xsqrt(1-1/x^2))/(xsqrt(4+1/x))$

$= sqrt(1-1/x^2)/sqrt(4+1/x)$

Evaluating the limit as $xrarroo$ , we get

$lim_(xrarroo)sqrt(x^2-1)/sqrt(4x^2+x) = sqrt(1-0)/sqrt(4+0) = 1/2$

How do you find the limit of (sqrt(x^2-1) ) / (sqrt(4x^2+x) )(sqrt(x^2-1) ) / (sqrt(4x^2+x) ) as x approaches infinity?