<s>[INST] Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$ 
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$

The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to 
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$. 
Hints and solutions are appreciated. Sorry if this is a duplicate.
 [/INST] $f(x):= \dfrac {x-1}{√x} -\dfrac{x-1}{\sqrt{x+1}}=$
$√x -\dfrac{1}{√x} -\sqrt{x+1} + \dfrac{2}{\sqrt{x+1}}$.
Note: 
$√x-\sqrt{x+1}= \dfrac{-1}{√x +\sqrt{x+1}}$ .
Combining:
$|f(x)| \le |\dfrac{-1}{√x+\sqrt{x+1}}| +|\dfrac{1}{√x}|$
$+ |\dfrac{2}{\sqrt{x+1}}|.$
$|f(x)| \lt \dfrac{1}{2√x} +\dfrac{1}{√x} + \dfrac{2}{2√x}.$
$0\le |f(x)|\lt \dfrac{5}{2√x}.$
$0\le \lim_{x \rightarrow \infty} |f(x)| \le$
$\lim_{x \rightarrow \infty} (5/2)\dfrac{1}{√x} = 0$.
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