Find where f is continuous and differentiable

**mremwo** · Mar 31st 2011, 08:27 AM

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

**TheEmptySet** · Mar 31st 2011, 08:37 AM

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 $\delta = \epsilon$

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

Now the right limit is where this get interesting!

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

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

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

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