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Question:

Calculate

$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$

without using L'Hospital's rule.

Attempted solution:

First we multiply with the conjugate expression:

$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$

Simplifying gives:

$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$

Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:

$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$

The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.

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There's an error in the factorisation of the denominator. No need to factor $x$ in the numerator, as you have a theorem for the limit at $\infty$ of a rational function. You should obtain, if $x>0$: $$\lim_{x \to+ \infty}\frac{3x-1}{x\Bigl(\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}\Bigr)} = \lim_{x \to +\infty} \frac{3x - 1}x\cdot\frac1{\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}}= 3\cdot\frac12.$$ Similarly the limit as $x\to-\infty\,$ is $\,-\dfrac32$.

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Your extraction of $x$ is incorrect: what you need is $$ \sqrt{x^2+3x} = \sqrt{x^2(1+3/x)} = x\sqrt{1+3/x}. $$ Then you have $$ \lim_{x \to \infty} \frac{3x-1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} = \lim_{x \to \infty} \frac{3}{\sqrt{1+3/x}+\sqrt{1+1/x^2}} - \frac{1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} $$ The square roots in this expression all tend to $1$, so the first term tends to $3/2$, the second to $0$.

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    $\begingroup$ youncan leave the numerator as $3-1/x$, which just converges to $3$. No reason to break it into two limits. $\endgroup$ Jun 13 '15 at 23:32
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I do not agree with your result after factorizing in the last step. I get

$$\frac{x(3-1/x)}{x\left(\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}\right)}=\frac{3-1/x}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}}\rightarrow 3/2$$

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$$ \sqrt{x^2+3x} - \sqrt{x^2+1} = x\sqrt{1+\frac{3}{x}} - x\sqrt{1+\frac{1}{x^2}} = x\left(\sqrt{1+\frac{3}{x}} -\sqrt{1+\frac{1}{x^2}} \right) $$ expand the radicals we find $$ \sqrt{1+\frac{3}{x}} = 1 + \frac{3}{2x} + O(x^{-2})\\ \sqrt{1+\frac{1}{x^2}} = 1 + O(x^{-2}) $$ put it all together $$ \lim_{x\to\infty}\left(\sqrt{x^2+3x} - \sqrt{x^2+1}\right) \to x\left(1 + \frac{3}{2x}-1\right) = \frac{3}{2} $$

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$$(\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) =(x \sqrt{1 + 3/x} - x\sqrt{1+ x^{-2} })$$

$$ =x\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) $$

$$ \left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) =1 +\frac{3}{2}\frac{1}{x} + O(x^{-2}) - (1+ O(x^{-2})) $$ using Taylor's theorem. So we get

$$ \frac{3}{2} + O(x^{-1}) $$

which converges to $3/2.$

Note that this approach is more easily generalizable to cases of different powers, eg cube roots.

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