How can we calculate this limit How can i calculate this limit :
$$\lim_{x \rightarrow 1} \frac{\sqrt[3]{x^2}-\sqrt[3]{(x-1)^2}-1}{x-1}$$
I cannot calculate this square root
Please help

Rewrite the expression as follows:
\begin{equation}
\lim_{x\to 1}\frac{x^{2/3}-1}{x-1} - \lim_{x\to 1} \frac{(x-1)^{2/3}}{x-1} \tag{*}
\end{equation}
The second limit is
$$\lim_{x\to 1^+} \frac{1}{(x-1)^{1/3}} =  \infty,$$
$$\lim_{x\to 1^-} \frac{1}{(x-1)^{1/3}} =  -\infty.$$
To simplify the first limit in (*), let $x^{1/3} -1 = t$, or $x = (t+1)^3$, then $x \to 1^+$ means $t \to 0^+$ and $x \to 1^-$ means $t \to 0^-$.  Then,
$$\lim_{x\to 1^+}\frac{x^{2/3}-1}{x-1} = \lim_{x\to 1^+}\frac{(x^{1/3}-1)(x^{1/3}+1)}{x-1} = \lim_{t \to 0^+}\frac{(t+1-1)(t+1+1)}{(t+1)^3-1} =  \lim_{t \to 0^+} \frac{t(t+2)}{t^3 + 3t^2 + 3t} = \lim_{t \to 0^+}\frac{t+2}{t^2 + 3t +3} = \frac{2}{3}. $$
It is easy to that
$$\lim_{x\to 1^-}\frac{x^{2/3}-1}{x-1} = \frac{2}{3}. $$
Hence, 
$$\lim_\limits{x\to 1^+}\frac{\sqrt[3]{x^2} -\sqrt[3]{(x-1)^2}-1}{x-1} = \frac{2}{3} - \infty = - \infty,$$
$$\lim_\limits{x\to 1^-}\frac{\sqrt[3]{x^2} -\sqrt[3]{(x-1)^2}-1}{x-1} = \frac{2}{3} - (-\infty) =  \infty.$$
