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 $1/4$? 

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 $-\sqrt 2$.
