f(x) = x^2 when x is rational, -x^2 when x is irrational. Does f'(0) exist?

[Grod28]

I'm having trouble with this problem: f(x) = x² when x is rational, -x² when x is irrational. Does f'(0) exist?

I think it doesn't exist because 0 is a rational number and left and right limits don't exist. Can anyone confirm or tell me if I'm wrong.

Any help will be truly appreciated!!!

Grod28

 

[Romsek]

I'm having trouble with this problem: f(x) = x² when x is rational, -x² when x is irrational. Does f'(0) exist?

I think it doesn't exist because 0 is a rational number and left and right limits don't exist. Can anyone confirm or tell me if I'm wrong.

Any help will be truly appreciated!!!

Grod28

use first principles

f'(x0) = lim(dx->0) (f(x0+dx) - f(x0))/dx

letting x0 = 0 here

f'(0) = lim(dx->0) (f(dx)-0)/dx = f(dx)/dx

so does f(dx)/dx have a limit as dx->0, if it does it's clearly going to be 0 as both 0^2 and -0^2 are 0.

use your epsilons and deltas to figure out whether f(dx)/dx has a limit at 0.

hint: the values of f in tiny neighborhoods about 0 do flip signs back and forth but they all tend towards 0 as the neighborhood shrinks to 0.

 

[Grod28]

use first principles

f'(x0) = lim(dx->0) (f(x0+dx) - f(x0))/dx

letting x0 = 0 here

f'(0) = lim(dx->0) (f(dx)-0)/dx = f(dx)/dx

so does f(dx)/dx have a limit as dx->0, if it does it's clearly going to be 0 as both 0^2 and -0^2 are 0.

use your epsilons and deltas to figure out whether f(dx)/dx has a limit at 0.

hint: the values of f in tiny neighborhoods about 0 do flip signs back and forth but they all tend towards 0 as the neighborhood shrinks to 0.

Thank you so much for your help!

G

 

[HallsofIvy]

left and right limits don't exist.

What did you do to check this before declaring they didn't exist?