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I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?

$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$

I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by $\sqrt{x^2}$, etc.

Obs.: Without using l'Hôpital's.

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Note that \begin{align} \sqrt{x^2+3x-4} - x & = \left(\sqrt{x^2+3x-4} - x \right) \times \dfrac{\sqrt{x^2+3x-4} + x}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{(\sqrt{x^2+3x-4} - x)(\sqrt{x^2+3x-4} + x)}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{x^2+3x-4-x^2}{\sqrt{x^2+3x-4} + x} = \dfrac{3x-4}{\sqrt{x^2+3x-4} + x}\\ & = \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1} \end{align} Now we get \begin{align} \lim_{x \to \infty}\sqrt{x^2+3x-4} - x & = \lim_{x \to \infty} \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}\\ & = \dfrac{3-\lim_{x \to \infty} 4/x}{1 + \lim_{x \to \infty} \sqrt{1+3/x-4/x^2} } = \dfrac{3}{1+1}\\ & = \dfrac32 \end{align}

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  • $\begingroup$ I'm really sorry for taking your time friend, I realized that I was working with $+1$ instead of $+x$ all the time. Thank you for your kind answer and sorry for this. $\endgroup$ – Thums Jun 9 '13 at 20:41
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Intuitively you can see this as follows:
Write $x^2+3x+4$ as $\left(x+\frac32\right)^2+\frac74$. For $x$ large this quantity is almost the same as $\left(x+\frac32\right)^2$. Therefore for $x$ large $\sqrt{x^2+3x+4}-x\sim\sqrt{\left(x+\frac32\right)^2}-x=\frac32$

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$$ \lim_{x \rightarrow \infty} \left(x^2 + 3x + 4\right)^{ \frac{1}{2}} - x $$ $$ = \lim_{x \rightarrow \infty} \left(x^2\left(1+\frac{3}{x}+\frac{4}{x^2}\right)\right)^{ \frac{1}{2}} - x $$ $$ = \lim_{x \rightarrow \infty} x\left(1+\frac{3}{x}+\frac{4}{x^2}\right)^{ \frac{1}{2}} - x $$

Then by Taylor expansion, we get that

$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{1}{2}\left(\frac{3}{x}+\frac{4}{x^2}\right)+\operatorname{o}\left(\frac{1}{x^2}\right)\right) - x $$

$$ = \lim_{x \rightarrow \infty} \frac{3}{2} + \operatorname{o}\left(\frac{1}{x}\right) = \frac{3}{2} $$

as required.

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