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 $1/4$? 
 A: 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.
A: Your first step of dividing both numerator and denominator by $x^2$ was a good idea. The denominator then converges to 4.
Now, you can write for the numerator:
$$ \frac{x\sqrt{x+1}(1-\sqrt{2x+3})}{x^2}= \frac{x\sqrt{x+1}}{x\sqrt{x}}\cdot\frac{1-\sqrt{2x+3}}{\sqrt{x}} $$
by splitting $x^2$ into $x\sqrt{x}$ and $\sqrt{x}$. 
The first term apparently goes to 1, while the second terms converges to $-\sqrt{2}$.
A: Set $x=1/h$ to get $$\lim_{h\to0^+}\dfrac{\sqrt{1+h}(h-\sqrt{2+3h})}{7h^2-6h+4}=\dfrac{\sqrt{1+0}(0-\sqrt{2+3\cdot0})}{7\cdot0^2-6\cdot0+4}=?$$
A: Dividing the numerator by $x\sqrt x\sqrt x$ and denominator by $x^2$, we have
$$ \lim_{x \to \infty} \frac{x \sqrt{x+1}(1-\sqrt{2x+3})}{7-6x+4x^2}
= \lim_{x \to \infty} \frac{\sqrt{1+\dfrac1x}\left(\dfrac1{\sqrt x}-\sqrt{2+\dfrac3{\sqrt x}}\right)}{\dfrac7{x^2}-\dfrac6x+4}=\frac{1\cdot(0-\sqrt2)}4.$$
A: The $x\sqrt{2/x+\cdots}$ term does not go to zero, but to infinity. Even when you multiply out the numerator and get a $\bigl(\sqrt{1/x+\cdots}\bigr)x\sqrt{2/x+\cdots}$ term, this goes to $\sqrt2$ rather than to $0$.
The term that arises from the $1$ when you multiply out does go to zero.
What you should probably do instead is rewrite the original numerator as
$$ x\sqrt{x+1}(1-\sqrt{2x+3}) = x^2\sqrt{1+1/x}\left(\sqrt{1/x}-\sqrt{2+3/x}\right) $$
where you can then reduce by $x^2$ and see the remaining factors go towards $1$ and $-\sqrt 2$.
