Computing a Limit + Justification

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Homework Statement



find the limit of [itex]\sqrt{x^2+x}[/itex]-[itex]\sqrt{x^2-x}[/itex] as x approaches infinity

Homework Equations





The Attempt at a Solution




Multiplying the original expression by
[itex]\frac{sqrt(x^2+x)+sqrt(x^2-x)}{sqrt(x^2+x)+sqrt(x^2-x)}[/itex]

I get the following:

[itex]\frac{2x}{sqrt(x^2+x)+sqrt(x^2-x)}[/itex]

I could use L'Hopital's rule here, but that just makes the expression more ugly and my professor recommended another way to solve it (but I've forgotten his recommendation!). The idea was something like this though:

We notice that the denominator looks a lot like [itex]\sqrt{x^2}[/itex]+[itex]\sqrt{x^2}[/itex] that is 2x suggesting the limit is 1. However, we have to deal with the other terms in the denominator to justify that answer.
 
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Heute said:

Homework Statement



find the limit of [itex]\sqrt{x^2+x}[/itex]-[itex]\sqrt{x^2-x}[/itex] as x approaches infinity

Homework Equations





The Attempt at a Solution




Multiplying the original expression by
[itex]\frac{sqrt(x^2+x)+sqrt(x^2-x)}{sqrt(x^2+x)+sqrt(x^2-x)}[/itex]

I get the following:

[itex]\frac{2x}{sqrt(x^2+x)+sqrt(x^2-x)}[/itex]

I could use L'Hopital's rule here, but that just makes the expression more ugly and my professor recommended another way to solve it (but I've forgotten his recommendation!). The idea was something like this though:

We notice that the denominator looks a lot like [itex]\sqrt{x^2}[/itex]+[itex]\sqrt{x^2}[/itex] that is 2x suggesting the limit is 1. However, we have to deal with the other terms in the denominator to justify that answer.
Factor the expressions in each of the radicals in the denominator like so:
x2(1 + 1/x) and x2(1 - 1/x)

Now bring the x2 factors out of the radicals and factor the resulting expression. You should be able to evaluate the limit then.
 
Ah! It's so obvious now! Thanks.