<s>[INST] 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 $0$
 [/INST] $$\displaystyle \lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$$
$$=-2\cdot\dfrac1{\lim_{x\to1}\{1+\cos(x-1)\}}\cdot\left(\lim_{x\to1}\dfrac{\sin(x-1)}{x-1}\right)^2\cdot\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
Set $\sqrt x-1=y\implies x=(y+1)^2$ in 
$$\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
to get $$\lim_{y\to0}\dfrac{\{(1+y)^2-1\}^2}{(1+y)^4-2(y+1)+1} =\lim_{y\to0}\dfrac{y^2(2+y)^2}{2y+O(y^2)}=\cdots=0$$
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