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.
 A: Using your work but multiplying by conjugate:
$$(x-1)\frac{\sqrt{x+1}-\sqrt x}{\sqrt x\sqrt{x+1}}=\frac{x-1}{\sqrt x\sqrt{x+1}\left(\sqrt x+\sqrt{x+1}\right)}\le\frac{2x}{x\cdot2\sqrt x}=\frac1{\sqrt x}\xrightarrow[x\to\infty]{}0$$
and since the expression is clearly positive, say for $\;x>10\;$ , then we're done with the squeeze theorem
A: Note that $\frac{x-1}{\sqrt{x}} = \sqrt{x} - \frac{1}{\sqrt{x}}$, for example, so the limit is equal to $$\lim_{x \to \infty} \sqrt{x} - \sqrt{x+1} - \frac{1}{\sqrt{x}} + \frac{2}{\sqrt{x+1}}$$
which by arithmetic of limits is $$\lim_{x \to \infty} \sqrt{x} - \sqrt{x+1}$$
which is well-known to be $0$.
A: You can use binomial approximation to get $$\sqrt{1+\frac 1x} \approx_{x \to \infty} 1+\frac 1{2x} $$
A: $$\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\frac{(x-1)\sqrt{x+1}-(x-1)\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}=\frac{x-1}{\sqrt{x}\sqrt{x+1}(\sqrt{x+1}+\sqrt{x})}\to0$$
A: Consider $$A=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}\sqrt{1+\frac 1x}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}}\left(1+\frac 1x \right)^{-1/2}$$ Now, by Taylor (or generalized binomial theorem
$$\left(1+\frac 1x \right)^{-1/2}=1-\frac{1}{2 x}+\frac{3}{8 x^2}-\frac{5}{16
   x^3}+O\left(\frac{1}{x^4}\right)$$ making
$$A=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}}\left(1-\frac{1}{2 x}+\frac{3}{8 x^2}-\frac{5}{16
   x^3}+O\left(\frac{1}{x^4}\right)\right)$$ Expand and simplify to get 
$$A=\frac 1{2 \sqrt x}-\frac 7 {8 x \sqrt x}+O\left(\frac{1}{x^{5/2}}\right)$$ which for sure shows the limit.
But it ia also a good approximation of the function value. Suppose $x=100$; the exact value would be $\frac{99}{10}-\frac{99}{\sqrt{101}}\approx 0.0491318$; the first term of the expansion would give $\frac 1{20}= 0.05$ while including the second term would give $\frac{393}{8000}=0.049125$.
A: \begin{align*}
f(x)&=x\cdot\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\right)+\dfrac{1}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x}}\\
&=\dfrac{\sqrt{x}}{\sqrt{x+1}}\dfrac{1}{\sqrt{x+1}+\sqrt{x}}-\dfrac{1}{\sqrt{x+1}\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+\sqrt{x+1}}\\
&\rightarrow 1\cdot 0-0\\
&=0.
\end{align*}
A: $$\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\frac{x-1}{\sqrt{x}}\left(1-\frac{\sqrt{x}}{\sqrt{x+1}}\right)=\frac{x-1}{\sqrt{x}}\left(1-\frac{1}{\sqrt{1+\frac1x}}\right)=\frac{x-1}{\sqrt{x}}\left(1-\left({1+\frac1x}\right)^{-\frac12}\right)=\frac{x-1}{\sqrt{x}}\left(1-1+\frac1{2x}+o\left(\frac1{x}\right)\right)=\frac{x-1}{\sqrt{x}}\left(\frac1{2x}+o\left(\frac1{x}\right)\right)=\frac{x-1}{x\sqrt{x}}\left(\frac1{2}+o(1)\right)\to0$$
A: $f(x):= \dfrac {x-1}{√x} -\dfrac{x-1}{\sqrt{x+1}}=$
$√x -\dfrac{1}{√x} -\sqrt{x+1} + \dfrac{2}{\sqrt{x+1}}$.
Note: 
$√x-\sqrt{x+1}= \dfrac{-1}{√x +\sqrt{x+1}}$ .
Combining:
$|f(x)| \le |\dfrac{-1}{√x+\sqrt{x+1}}| +|\dfrac{1}{√x}|$
$+ |\dfrac{2}{\sqrt{x+1}}|.$
$|f(x)| \lt \dfrac{1}{2√x} +\dfrac{1}{√x} + \dfrac{2}{2√x}.$
$0\le |f(x)|\lt \dfrac{5}{2√x}.$
$0\le \lim_{x \rightarrow \infty} |f(x)| \le$
$\lim_{x \rightarrow \infty} (5/2)\dfrac{1}{√x} = 0$.
