How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule? 
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$

Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is $0$

We start by using the double-angle identity for the cosine function to write
$$2\cos (x-1)-2=-4\sin^2\left(\frac{x-1}{2}\right) \tag 1$$
Next, we factor the denominator of the term of interest to find
$$\begin{align}
x^2-2x^{1/2}+1&=(x^{1/2}-1)(x^{3/2}+x+x^{1/2}-1)\\\\
&=\frac{(x^{3/2}+x+x^{1/2}-1)(x-1)}{x^{1/2}+1} \tag 2
\end{align}$$
Then, using $(1)$ and $(2)$ we have
$$\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}=\left(-4\frac{x^{1/2}+1}{x^{3/2}+x+x^{1/2}-1}\right) \times \left(\frac{\sin\left(\frac{x-1}{2}\right)}{x-1}\right)\times\left( \sin\left(\frac{x-1}{2}\right) \right)\tag 3$$
The first parenthetical term on the right-hand side of $(3)$ approaches $-4$ as $x\to 1$, the second term approaches $1/2$, and the third term approaches $0$.
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}\left(\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}\right)=0}$$
