Limit as x tends to infinity, without Laurent

[Skeptic.]

Homework Statement

I want to find the following limit, $\lim_{x \rightarrow \infty } x( \sqrt{ x^{2} +9} -x)$, without using the Laurent series

Homework Equations

None.

The Attempt at a Solution

I used the Laurent Series to expand the square root, giving $x((x+\frac{9}{2x})-x)$, then giving the limit as $\frac{9}{2}$ . How would one go about this question without using the above method?

 

[Simon Bridge]

lim_x→∞x(√ x²+9−x)

... to me that reads:$$\lim_{x\to\infty}x\sqrt{x^2-x+9}$$ ... I don't think that converges: perhaps you meant to write something different.

Mostly someone would deal with a limit like that by remembering how the functions work for the limit in question.
i.e. as x gets very big, large powers of x will come to dominate a sum.

So: $$\lim_{x\to\infty} \frac{x}{\sqrt{x^4-2x+9000}}=\lim_{x\to\infty}\frac{1}{x} = 0$$ ... which you can check on a calculator.

I find it suggestive that the 9 under the root is a square number - so if you want to be more rigorous-ish, try completing the square, look for a substitution maybe?

 

[Mark44]

Homework Statement

I want to find the following limit, $\lim_{x \rightarrow \infty } x( \sqrt{ x^{2} +9} -x)$, without using the Laurent series

Homework Equations

None.

The Attempt at a Solution

I used the Laurent Series to expand the square root, giving $x((x+\frac{9}{2x})-x)$, then giving the limit as $\frac{9}{2}$ . How would one go about this question without using the above method?

... to me that reads:$$\lim_{x\to\infty}x\sqrt{x^2-x+9}$$ ... I don't think that converges: perhaps you meant to write something different.

The OP has something different. I don't believe it was edited, but perhaps it was.

@Skeptic, multiply by $\sqrt{x^2 + 9} + x$ over itself, and you'll be able to take the limit.

 

[Simon Bridge]

Oh I see - my display is not showing the line over the square-root in post #1, and it didn't drag to quote in my reply either - but I do see it quoted in post #3.
Cheers.