A strange thing when evaluating a basic limit I have attempted to evaluate the limit $\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)$. I understand, intuitively, that the limit is equal to zero, but when trying to show it, something strange happened. I've tried to evaluate the limit based on a technique that uses the fact that $\lim _{x\to \infty }\left(\frac{1}{x^n}\right)\:=\:0,\:n>0$, and divided both the numerator and the denominator of the fraction by $x$:
\begin{align}
& \lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{x}}\right) =\lim_{x\to \:\infty} \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{\sqrt{x^2}}}\right) =\lim_{x\to \:\:\infty \:\:}\left(\frac{\frac{1}{x}}{\sqrt{\frac{x-1}{x^2}}}\right) \\[10pt]
= {} & \lim_{x\to \:\:\:\infty \:\:\:} \left(\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}\right)=\frac{\frac 1 \infty}{\sqrt{\frac 1 \infty -\frac{1}{\infty ^2}}}=\frac{0}{\sqrt{0-0}}=\frac{0}{0}= \text{indeterminate}
\end{align}
Moreover, when I divided the numerator and the denominator by $\sqrt{x}$, I've actually gotten the correct result:
$$\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{\sqrt{x}}}{\frac{\sqrt{x-1}}{\sqrt{x}}}\right)=\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{\frac{x-1}{x}}}\right)=\lim_{x\to \:\infty \:}\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{1-\frac{1}{x}}} \right) = \frac{\frac{1}{\sqrt{\infty }}}{\sqrt{1-\frac{1}{\infty }}}=\frac{0}{\sqrt{1-0}} = \frac{0}{1}=0$$ 
Why is there a difference in the results? Why does dividing by $\sqrt{x}$ is allowed, but not by dividing by $x$, or have I done an arithmetic/algebraic mistake that caused the result to be wrong? 
Help will be much appreciated :)

Yes for that limit we have the indeterminate form
$$\lim_{x\to\infty} \frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\frac{0}{0}$$
but we can eliminate the indetermination by
$$\lim_{x\to\infty}\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\lim_{x\to\infty}\frac{\sqrt{\frac1x} }{\sqrt{\frac1x}}\frac{\sqrt{\frac1x}}{\sqrt{1-\frac{1}{x}}}=\frac{0}1=0$$
therefore it seems there is nothing special to that.
The latter manipuilation explain also the reason for a different result in the second case when we divide numerator and denominator by $\sqrt x$.
