# prove a derviative does not exist at a point

Printable View

• Apr 21st 2013, 04:00 AM
amyw
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.
• Apr 21st 2013, 04:10 AM
Plato
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.
• Apr 22nd 2013, 01:36 PM
amyw
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?
• Apr 22nd 2013, 01:52 PM
Plato
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?
• Apr 22nd 2013, 01:54 PM
Bradyns
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. :)
• Apr 22nd 2013, 03:11 PM
amyw
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.
• Apr 22nd 2013, 03:20 PM
amyw
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
• Apr 22nd 2013, 03:32 PM
Plato
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.
• Apr 22nd 2013, 03:39 PM
amyw
Re: prove a derviative does not exist at a point
Of course. Duh...