Finding the Limit in: $\lim\limits_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$ Need some help finding this limit:
$$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$$
Here is what I have so far:
$$\lim_{x\rightarrow1}\dfrac{\dfrac{1-\sqrt{x}}{\sqrt{x}}}{x-1}$$ 
$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{x-1}$$
$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}$$
At this point I keep getting results I don't like, I have tried multiplying by the conjugate but I keep getting denominators of $0$. What am I missing here?
Thanks

Another approach is to make the change of variables $y=\sqrt{x}$; because this is a continuous function, we know that the limit as $x\rightarrow 1$ is the same as the limit as $y\rightarrow 1$, so that we have $$\lim_{x\rightarrow 1}{ \frac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}}= \lim_{y\rightarrow 1}{\frac{1-y}{y^3-y}}=\lim_{y\rightarrow 1}{\frac{-(y-1)}{y(y-1)(y+1)}}=\lim_{y\rightarrow 1}{\frac{-1}{y(y+1)}}=-\frac{1}{2}$$
