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Trying to work out a limit from a past exam in Calculus and Linear Algebra 1, now I know the answer is 0, and I have a worked solution from Wolframalpha is, but the answer is definitely not done the way they did it(computationally) as it was some 20 steps long.

My thought is, I take it to be a fraction somehow, and I divide by highest power, or perhaps apply the conjugate of the function.

$\lim_{{x}\to{\infty}}$ $x-\sqrt{1+x^2}$

I did the conjugate and got it down to $${-}\frac{1}{x+\sqrt{1+x^2}}$$ Which I can now rationalize to be 0, as dividing by the highest power will definitely give me zero, and I am pretty sure taking the highest power on $\sqrt{1+x^2}$ just gives me 1, which gives me 1+1 on the bottom(Which means the limit does exist).

Edit: I want to solve this without L'Hop if possible for the record.

How does one work this limit out? Am I on the right track?

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HINT: $$ \frac{1}{x + \sqrt{1 + x^2}} = \frac{\frac{1}{x}}{1 + \sqrt{\frac{1}{x^2} + 1}} $$ for $x > 0$. Now what happens when $x \to \infty$?

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I don't believe $$\frac{{\sqrt{1+x^2}}}{x} = \sqrt{1+{\frac{1}{x^2}}}$$ –  Display Name Oct 18 '13 at 1:53
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When $x > 0$,$$\frac{\sqrt{1 + x^2}}{x} = \frac{\sqrt{1 + x^2}}{\sqrt{x^2}} = \sqrt{\frac{1}{x^2} + 1}$$ –  tylerc0816 Oct 18 '13 at 1:57
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For $x\to\infty$

$$\sqrt{1+x^2} = O(x)$$

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If you take what you got by conjugating, factor an $x^2$ out of what you have in the square root. Then see what simplifying you can do; you should be able to find the limit after doing the simplification.

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