Finding horizontal tangent points using the limit definition of the derivative

First expand $f(x+h) = (x+h)^4 - 2(x+h)^2 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 - 2x^2 - 4xh - 2h^2$. Subtracting $f(x) = x^4 - 2x^2$ leaves $4x^3h + 6x^2h^2 + 4xh^3 + h^4 - 4xh - 2h^2$. Dividing by $h$ gives $\frac{f(x+h)-f(x)}{h} = 4x^3 + 6x^2h + 4xh^2 + h^3 - 4x - 2h$.

Taking the limit as $h \to 0$ eliminates every term containing $h$, leaving $f'(x) = 4x^3 - 4x$. Setting this equal to zero: $4x^3 - 4x = 0 \Rightarrow 4x(x^2-1)=0 \Rightarrow x(x-1)(x+1)=0$, so $x = 0, 1, -1$. These are the points where the tangent line is horizontal.