Find $\displaystyle\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$ using squeeze theorem 
Find $$\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$$ using squeeze theorem

Tried using binomial expansion, but have no idea on how to continue.
 A: From
$$(x+1)(x+2)<\left(x+\tfrac32\right)^2,$$ you draw (using the conjugate, on the right)
$$-\frac32=x-\left(x+\tfrac32\right)<x-\sqrt{(x+1)(x+2)}<\frac{x^2-(x+1)(x+2)}{x+\left(x+\frac32\right)}=-\frac32\frac{x+\frac23}{x+\frac34}.$$

A: $$\begin{align*}&x-\sqrt{x+1}\sqrt{x+2}=\frac{x^2-(x+1)(x+2)}{x+\sqrt{x+1}\sqrt{x+2}}=\frac{-3x+2}{x+\sqrt{x^2+3x+2}}=\\{}\\
&\frac{-3+\frac2x}{1+\sqrt{1+\frac3x+\frac2{x^2}}}\xrightarrow[x\to\infty]{}\frac{-3+0}{1+\sqrt{1+0+0}}=-\frac32\end{align*}$$
A: Your idea of a use of the binomial theorem was good, writing$$A=x - \sqrt{x+1}\times\sqrt{x+2}=x-\sqrt x \times\sqrt{1+\frac{1}{x}}\times\sqrt x \times\sqrt{1+\frac{2}{x}}$$ that is to say $$A=x\left(1-\sqrt{1+\frac{1}{x}}\times \sqrt{1+\frac{2}{x}}\right)$$ Now consider, using the binomial theorem with $t=\frac 1x$$$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+\cdots$$ $$\sqrt{1+2t}=1+t-\frac{t^2}{2}+\cdots$$ $$\sqrt{1+t}\times\sqrt{1+2t}=1+\frac{3 t}{2}-\frac{t^2}{8}+\cdots$$ Replace $t$ by $\frac 1x$ to get $$A=x\left(1-1-\frac{3 }{2x}+\frac{1}{8x^2}+\cdots\right)=-\frac{3 }{2}+\frac{1}{8x}+\cdots$$
