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 $0$

Set $x-1=u$ and rewrite the expression:


*

*$2\cos(x-1)-2=2\cos u-2=-u^2+o(u^2)$,

*$\begin{aligned}[t]x^2-2\sqrt x+1&=(1+u)^2-2\sqrt{1+u}+1=1+2u+u^2-2-u+\dfrac{u^2}4+o(u^2)+1\\&=u+\dfrac{5u^2}4+o(u^2)
\end{aligned}.$


Thus $2\cos(x-1)-2\sim_1 -(x-1)^2$, $\;x^2-2\sqrt x+1\sim_1 x-1$, and
$$\dfrac{2\cos(x-1)-2}{x^2-2\sqrt x+1}\sim_1\frac{-(x-1)^2}{x-1}=1-x \xrightarrow[x\to 1]{} 0.$$
