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 $0$
 A: 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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}\left(\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}\right)=0}$$
A: 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
$$\dfrac{2\cos(x-1)-2}{x^2-2\sqrt x+1}\sim_1\frac{-(x-1)^2}{x-1}=1-x \xrightarrow[x\to 1]{} 0.$$
A: Let $f(x) = 2 \cos (x-1), g(x) = x^2 - 2 \sqrt x.$ We are looking at
$$\frac{f(x) - f(1)}{g(x) -g(1)} = \frac{(f(x) - f(1))/(x-1)}{(g(x) -g(1))/(x-1)}.$$
By the definition of the derivative, the above $\to f'(1)/g'(1)$ as $x\to 1.$ This is a simple computation.
A: $$\displaystyle \lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$$
$$=-2\cdot\dfrac1{\lim_{x\to1}\{1+\cos(x-1)\}}\cdot\left(\lim_{x\to1}\dfrac{\sin(x-1)}{x-1}\right)^2\cdot\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
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 $$\lim_{y\to0}\dfrac{\{(1+y)^2-1\}^2}{(1+y)^4-2(y+1)+1} =\lim_{y\to0}\dfrac{y^2(2+y)^2}{2y+O(y^2)}=\cdots=0$$
A: Set $u=\sqrt x$, then
$$...=\lim_{u\to 1}\frac{2\cos(u^2-1)-2}{u^4-2u+1}=\lim_{u\to 1}\frac{2\cos(u^2-1)-2}{(u-1)(u^3+u^2+u-1)}.$$
And use the fact that $$\cos(y-1)=1-\frac{(y-1)^2}{2}+o((y-1)^2).$$
