Differentiability and its relation with continuity

To study the differentiability at $x=1$ we must do the side derivatives at $x=1$ and see if they exist and coincide.

a) $f(x)=x^3+x-2$

$$f'(0^-)=\lim_{\Delta x \to 0^-}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{(1+\Delta x)^3+(1+\Delta x)-2-1^3-1+2}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^-}\frac{1+3\Delta x+3\Delta x^2+\Delta x^3+1+\Delta x-2}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{\Delta x^3+3\Delta x^2+4\Delta x}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^-}(\Delta x^2+3\Delta x+4)=4$$

$$f'(0^+)=\lim_{\Delta x \to 0^+}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{(1+\Delta x)^3+(1+\Delta x)-2-1^3-1+2}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^+}\frac{1+3\Delta x+3\Delta x^2+\Delta x^3+1+\Delta x-2}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{\Delta x^3+3\Delta x^2+4\Delta x}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^+}(\Delta x^2+3\Delta x+4)=4$$

The two values coincide, both approaching from the right and from the left. Therefore, this function is differentiable at $x=1$.

As it is differentiable we can say that it is continuous.

b) $f(x)=\dfrac{1}{(x-1)^2}$

$$f'(0^-)=\lim_{\Delta x \to 0^-}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{\dfrac{1}{(1+\Delta x-1)^2}-\dfrac{1}{(1-1)^2}}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^-}(\dfrac{\Delta x}{\Delta x^2}-\dfrac{\Delta x}{0})=-\infty$$

$$f'(0^+)=\lim_{\Delta x \to 0^+}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{\dfrac{1}{(1+\Delta x-1)^2}-\dfrac{1}{(1-1)^2}}{\Delta x}=$$ $$=\lim_{\Delta x \to 0^+}(\dfrac{\Delta x}{\Delta x^2}-\dfrac{\Delta x}{0})=+\infty$$

The two values do not coincide, thus the function is not differentiable at $x=1$.

We can't say anything about the continuity in $x=1$.