<s>[INST] Evaluate this limit at infinity $\lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}$ Problem: Find the limit of \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
\end{align*}
Attempt at solution. The back of my textbook gives the answer as $-\frac{1}{4} \sqrt{2}$. Here's what I did: \begin{align*} \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2} \\ = \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2 (4-6/x+7/x^2)} \\ \\ = \lim_{x \to \infty} \frac{\sqrt{x^2(1/x+1/x^2)}(1-\sqrt{x^2(2/x+3/x^2)}}{x(4-6/x+7/x^2)} \\ \\ \lim_{x \to \infty} \frac{\sqrt{1/x+1/x^2}(1-x \sqrt{2/x+3/x^2})}{4-6/x+7/x^2} 
\end{align*} If I now evaluate this limit, everything in the numerator goes to zero except $1$. And the denominator leaves me with $4$. So I thought the answer should be $1/4$? 
 [/INST] You can divide the numerator and denominator by $x^2$ and distribute in an appropriate manner to obtain:
$$\begin{align*}\frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}& = \frac{\frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{x^2}}{\frac{7-6x+4x^2}{x^2}}\\\\& =\frac{\frac{x}{x}\cdot \frac{\sqrt{x+1}}{\sqrt{x}}\cdot\left(\frac{1-\sqrt{2x+3}}{\sqrt{x}}\right)}{\frac{7-6x+4x^2}{x^2}}\\\\ & = \frac{\sqrt{\frac{x+1}{x}}\cdot\left(\frac{1}{\sqrt{x}}-\sqrt{\frac{2x+3}{x}}\right)}{\frac{7-6x+4x^2}{x^2}} =\frac{\sqrt{1+\frac{1}{x}}\cdot\left(\frac{1}{\sqrt{x}}-\sqrt{2+\frac{3}{x}}\right)}{\frac{7}{x^2}+\frac{6}{x}+4}\end{align*}$$
This technique is usually performed when both, the numerator and denominator are polynomials, however, it can be still performed here and it becomes clearer if you think the numerator as a $1+\frac{1}{2}+\frac{1}{2}=2$ degree 'polynomial', in some sense.
At this point you can take the limit.
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