<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] Set $x=1/h$ to get $$\lim_{h\to0^+}\dfrac{\sqrt{1+h}(h-\sqrt{2+3h})}{7h^2-6h+4}=\dfrac{\sqrt{1+0}(0-\sqrt{2+3\cdot0})}{7\cdot0^2-6\cdot0+4}=?$$
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