Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of $ x$ but do not possess a derivative for any value. We now consider the celebrated function given by Weierstrass to show this fact. It will be shown that if $ f(x)=

# G.H. Hardy

Consider a sequence of functions as follows:- $ f_1 (x) = \sqrt {1+\sqrt {x} } $ $ f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } } $ $ f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } } $ ……and so on to $ f_n