prove a derviative does not exist at a point

Hi All,

I was asked to prove that the derivative at x=0 of f(x)=x^2/3 does not exist. I know that it doesn't because the function is a corner at x=0, but I'm not sure how to start. I know the definition of derivative and I believe I need to do this by contradiction using the definition and cases, but what do I do now?(Crying)

Thanks.

Re: prove a derviative does not exist at a point

Quote:

Originally Posted by

**amyw** I was asked to prove that the derivative at x=0 of f(x)=x^2/3 does not exist. I know that it doesn't because the function is a corner at x=0, but I'm not sure how to start.

$\displaystyle \frac{{f(0 + h) - f(0)}}{h} = \frac{1}{{\sqrt[3]{h}}}$ now show that $\displaystyle {\lim _{h \to 0}}\frac{1}{{\sqrt[3]{h}}}$ does no exist.

Re: prove a derviative does not exist at a point

I have only proved that a limit is a real number. I can see that the limit as h-->0 from the left is negative infinity and as h-->0 from the right the limit is positive infinity. Since there isn't a unique limit then the limit DNE. But, I don't know how to show a limit is infinity positive or negative. Help?

Re: prove a derviative does not exist at a point

Quote:

Originally Posted by

**amyw** I have only proved that a limit is a real number. I can see that the limit as h-->0 from the left is negative infinity and as h-->0 from the right the limit is positive infinity. Since there isn't a unique limit then the limit DNE. But, I don't know how to show a limit is infinity positive or negative. Help?

To show that $\displaystyle {\lim _{x \to \0^+ }}f(x) = \infty $ you must show

that if $\displaystyle N>0$ then $\displaystyle \exist \delta>0$ such that $\displaystyle 0<x<\delta \Rightarrow \;f(x) \ge N$.

In this case let $\displaystyle \delta=\frac{1}{N^3}$. Can you finish?

Re: prove a derviative does not exist at a point

If the limit does not exist

i.e. Limit from the postive =/= limit from the negative, then the function isn't continuous.

For a funtion to be differentiable at a point, it needs to be continuous at that same point, thus the limit needs to exist at that point.

You've shown that it isn't continuous at that point, then by definition it cannot be differentiate at that point. :)

Re: prove a derviative does not exist at a point

Okay, I figured this out. So when I show there is a delta so that 0<x<delta, and let delta equal 1/N^3, then N</= 1/3rooth, that means that the limit is positive infinity?

What about the left sided limit that goes to negative infinity? How do I do that?

Isn't that bit instrumental in showing that at 0 there are two different limits and therefor the limit doesn't exist at zero? And since the limit doesn't exist the original function has no derivative at x=0?

Thanks for the help Plato.

Re: prove a derviative does not exist at a point

No. I feel like I'm missing something crucial. I can see, even though I can't do it, that this takes care of the right sided limit. What about the left sided limit?

Thanks

Re: prove a derviative does not exist at a point

Quote:

Originally Posted by

**amyw** No. I feel like I'm missing something crucial. I can see, even though I can't do it, that this takes care of the right sided limit. What about the left sided limit?

The point is, I have shown the right-hand limit is not finite. Therefore the derivative at $\displaystyle x=0$ cannot exist.

That is all you are asked to do.

Re: prove a derviative does not exist at a point