Thread: Find where f is continuous and differentiable

For all n in N (natural numbers), let f(x):= x^n if x >= (gr/eq to) 0 and f(x)=0 for x<0. For which values of n is f' continuous at 0? Which values of n is f' differentiable at 0? I can understand why this is, but I don't know how to prove it using limit definitions. Thanks for your help

Originally Posted by mremwo

For all n in N (natural numbers), let f(x):= x^n if x >= (gr/eq to) 0 and f(x)=0 for x<0. For which values of n is f' continuous at 0? Which values of n is f' differentiable at 0? I can understand why this is, but I don't know how to prove it using limit definitions. Thanks for your help

You need to check the limit from both sides

The left one is easy. let $\displaystyle \delta = \epsilon$

Then $\displaystyle 0-x<\delta \implies \frac{f(x)-0}{x-0} =0 < \epsilon$

Now the right limit is where this get interesting!

$\displaystyle x-0 < \delta $ now we need to analyze

$\displaystyle \displaystyle \frac{x^n-0}{x-0}=x^{n-1}$

This need to equal the limit form the left! So what must the values of $\displaystyle n$ be?