Let I be an open interval that contains 0 and
f: from I to R. If there exist an alpha greater than 1 such that
for all x, prove that f is differentiable at x=0.
What happens when alpha = 1?
Please check if my proof is correct and fix the graph.
To prove that f(x) is differentiable, we need to prove that the following limit exists
By the limit theorems, the limit of the difference, is the difference of the limits.
In particular a=0, then is enough to check that that the limit of f(x)/x and
f(0)/x exist when x goes to 0.
By the comparison theorem for functions, the limit of the absolute value of x to the alpha goes to 0. Therefore, the limit of f(x) also goes to 0. Additionaly, for the squeeze theorem for functions f(0)=0. Therefore, the derivative exists and equals to 0.
If alpha is equal to 1, we have in the right side of the inequality the absolute value function that is continuous but not differentiable at x=0. As the derivative of the absolute value function is 1 when x is greater than 0, and -1 when x is less than 0, we would not be able to know if f(x) is differentiable at x=0.