MATH2060A Solution to Assignment 1
examine the di erentiability of j at 0 using de niton. Indeed, using the fact the cosine function is even, lim h!0 cosjhj cos0 h 0 = lim h!0 cosh 1 h = 0; from which we conclude that jis also di erentiable at x= 0. Hence jis di erentiable everywhere. A shortcut is to realize that the cosine is an even function, so j(x) = cosxis dif-ferentiable ...
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