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.

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 $\frac{393}{8000}=0.049125$.
