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Let $f$ be a function such that $x^2f(x)\geq x^2+x+1,\forall x\in\mathbb{R^*}$. Find the value of: $$\lim_\limits{x\to 0}{\left(\sqrt{f^2(x)+2f(x)+3}-f(x)\right)}$$

I think that we may need to apply a sandwich theorem for the limit, so let $g(x)=\sqrt{f^2(x)+2f(x)+3}-f(x)$. It is easy to prove that $g(x)>1$ near/close to 0, since $f(x)>0$ from the original relation. However, I cannot find any function $h$, such that $g(x)<h(x)$ near/close to 0 with $\lim_\limits{x\to 0}{h(x)}=1$. Any hint?

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    $\begingroup$ When $x\to0$, $f(x)\to+\infty$, hence the fact that $$\sqrt{f^2(x)+2f(x)+3}-f(x)=\frac{2f(x)+3}{\sqrt{f^2(x)+2f(x)+3}+f(x)}=\frac{2f(x)+o(f(x))}{f(x)+o(f(x))+f(x)}$$ shows the limit exists and is... $\endgroup$ – Did Sep 13 '15 at 15:52
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    $\begingroup$ $$\sqrt{f^2+2f+3} - f = 1 + \left(\sqrt{f^2+2f+3} - (f+1)\right) = 1 + \frac{2}{\sqrt{f^2+f+3}+(f+1)}$$ $\endgroup$ – achille hui Sep 13 '15 at 15:55
  • $\begingroup$ @Did I fully understand your approach (the limit is 1) but we haven't "officially" learnt asymptotic notation yet... Is there any other approach? $\endgroup$ – Jason Sep 13 '15 at 15:55
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    $\begingroup$ Sure: the very same without the asymptotic notation. For example, $$f(x)+1\leqslant\sqrt{f^2(x)+2f(x)+3}\leqslant f(x)+2.$$ $\endgroup$ – Did Sep 13 '15 at 16:21
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If you want to use the sandwich theorem, observe that whenever $x$ is positive, $f(x)$ also is positive and \begin{align} \left(f(x) + 1 + \frac{1}{f(x)}\right)^2 & = (f(x))^2 + 2f(x) + 3 + \frac{2}{f(x)} + \frac{1}{(f(x))^2} \\ & > (f(x))^2 + 2f(x) + 3 \\ & > \left(f(x) + 1\right)^2 \\ & > 0. \end{align}

Therefore $$f(x) + 1 < \sqrt{(f(x))^2 + 2f(x) + 3} < f(x) + 1 + \frac{1}{f(x)}$$ and $$1 < \sqrt{(f(x))^2 + 2f(x) + 3} - f(x) < 1 + \frac{1}{f(x)}.$$

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$f(x)\ge1+\frac 1x+\frac1{x^2} \Rightarrow$ as $x \to 0$ we get $f(x) \to \infty$.

Now multiplying and dividing by $\sqrt {f^2(x)+2f(x)+3} +f(x)$ we get,

$\frac {2f(x)+3}{\sqrt {f^2(x)+2f(x)+3}+f(x)} \Rightarrow$ $\frac {2+\frac 3{f(x)}}{\sqrt {1+\frac 2{f(x)}+\frac 3{f^2(x)}}+1}$.

Now as $x \to 0$, we get $\frac {2+0}{\sqrt 1 + 1} = 1$.

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You can still use the setup provided by @Did. Simply write

$$\sqrt{f^2+2f+3}-f=\frac{2f+3}{\sqrt{f^2+2f+3}+f}=\frac{2+3/f}{1+\sqrt{1+\frac{2}{f}+\frac{3}{f^2}}}$$

and then let $f\to \infty$

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