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
 A: I find the route via L'Hôpital to be shortest here.  
$$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}=
\lim_{x\rightarrow1}\frac{\frac{d}{dx}(\frac{1}{\sqrt{x}}-1)}{\frac{d}{dx}(x-1)}=
\lim_{x\rightarrow1}\;-\frac{1}{2}x^{-3/2}=-\frac{1}{2}$$  
Same result. Different path.  
A graph of the function shows the limit point at $-\frac{1}{2}$ as calculated. I would like to point out that the function is undefined at this point, but the graph fails to show this.  

A: $$\lim_{x\rightarrow 1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}= \lim_{x\rightarrow 1}\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}\cdot\frac{1+\sqrt{x}}{1+\sqrt{x}}=\lim_{x\rightarrow 1}\frac{1-x}{(x-1)\sqrt{x}(1+\sqrt{x})}=$$
$$=\lim_{x\rightarrow 1}-\frac{x-1}{\sqrt{x}(x-1)(1+\sqrt{x})}=\lim_{x\rightarrow 1}-\frac{1}{\sqrt{x}(1+\sqrt{x})}=-\frac{1}{2}. $$
A: Hint: $1-x=\big(1-\sqrt x\big)\big(1+\sqrt x\big)$.
A: 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 $$\lim_{x\rightarrow 1}{ \frac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}}= \lim_{y\rightarrow 1}{\frac{1-y}{y^3-y}}=\lim_{y\rightarrow 1}{\frac{-(y-1)}{y(y-1)(y+1)}}=\lim_{y\rightarrow 1}{\frac{-1}{y(y+1)}}=-\frac{1}{2}$$
