As you wrote, . Using Maclaurin series for , we get . Therefore, . This expression tends to when and to when , so there is no limit.1) Use the definition of derivative to show that function y = g(x) implicitly defined by equation: is not differentiable at point (0,1)

It also helps to have a sketch of the graph of y(x). This function is symmetric w.r.t. the vertical x = 0 axis because x occurs squared. Also, the behavior of near 0 is similar to that of , in particular, when . Therefore, has a sharp angle at (0,1), which is its maximum point. I believe that raising everything to power 3/2 does not fundamentally change this behavior.

This is pretty straightforward using the formula for the derivative of a fraction and the fact that .3) Let , Show that