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How might one go about taking the following limit without using L'Hopital's rule? I am stumped:

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

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    @Marvis: Hmm... That abstract duplicate feels a bit *too* abstract, since none of the answers there cover the simple way of dealing with the case $n=2$ (at least not obviously).2012-10-28

9 Answers 9

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First, factor out $x$ from the radical:

$\lim_{x \to \infty} x \left(\sqrt{1+\frac{1}{x}} - 1 \right)$

You can then Taylor expand the radical, as $1/x$ is small as $x$ becomes large. The relevant expansion is $(1+1/x)^{1/2} \approx 1 + \frac{1}{2x}$. This should make the result of the limit apparent.

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HINT

$\sqrt{x^2+x} - x = \dfrac{\left(\sqrt{x^2+x} - x\right) \left(\sqrt{x^2+x} + x \right)}{\sqrt{x^2+x} + x} = \dfrac{x^2+x - x^2}{\sqrt{x^2+x} + x}$

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$\sqrt{x^2+x}-x=\sqrt{x^2\left(1+\frac1x\right)}-x=x\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right)\;;$

let $u=\dfrac1x$, and this becomes $\frac{\sqrt{1+u}-1}u=\frac{\sqrt{1+u}-1}u\cdot\frac{\sqrt{1+u}+1}{\sqrt{1+u}+1}=\frac{u}{u\left(\sqrt{1+u}+1\right)}=\frac1{\sqrt{1+u}+1}\;.$

As $x\to\infty$, $u\to0^+$, so just calculate $\lim_{u\to0^+}\frac1{\sqrt{1+u}+1}\;.$

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    @providence: You mean if you couldn’t appeal to the fact that $f(u)=\sqrt{1+u}$ is continuous at $0$ to say that its limit as $u\to0$ is $1$? You’d probably essentially have to prove it from the the definition of limit.2012-10-29
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Hint:

Multiply and divide function under limit by $ \sqrt{x^2 + x} + x$: $\lim\limits_{x \to \infty}\left( \sqrt{x^2 + x} - x\right)=\lim\limits_{x \to \infty}\dfrac{\left( \sqrt{x^2 + x} - x\right)\left( \sqrt{x^2 + x} + x\right)}{ \sqrt{x^2 + x} + x}$

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$\lim_{x\to\infty} \sqrt{x^2+x}-x=\lim_{x\to\infty} (\sqrt{x^2+x}-x)\cdot \frac{ \sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty} \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\dots$

Can you take it from here?

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Observing that for $x>0$ we have $\sqrt{x^2+x}-x=|x|\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right),$ then using $(a-b)(a+b)=a^2-b^2$ gives us $\left(\sqrt{x^2+x}-x\right)\cdot\left(\sqrt{1+\frac1x}+1\right)=x\left(1+\frac1x-1\right)=1.$ Thus, $\sqrt{x^2+x}-x=\frac1{\sqrt{1+\frac1x}+1},$ which makes evaluation of the limit as $x\to\infty$ much simpler.

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You could complete the square inside the square root to give:

$\sqrt{\left(x+\frac 1 2\right)^2-\frac 1 4} - x=\left(x+\frac 1 2\right)\sqrt{1-\frac 1 {(2x+1)^2}}-x$

And use the binomial series expansion of the square root. I mention this because it is a different idea from that in the other answers - it is not necessarily a better one. But if you are taking a factor outside the square root, I think completing the square gives you a better chance of getting the right factor. This puts a square in the denominator, rather than a linear denominator as in some of the other suggestions.

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Hint: make a fraction of it and use $(a+b)(a-b)$.

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Second hint: rewrite the fraction you've made (per Berci's comment) into $\frac{1}{\sqrt{1+\frac{1}{x}}+1}.$