# Math Help - more differntiable

1. ## more differntiable

let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)
B) show that f is differentiable at x = 0 and show that f(0)=0
C) show that f' is not continuous at x = 0

2. Originally Posted by learn18
let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

A) use the chain rule and the product rule to show that f is differentiable at each c is not equal to 0 and find f'(c)
at x = c not= 0, f(x) is continuous and hence differentiable.

so f(c) = (c^2)sin(1/c)
=> f ' (c) = 2csin(1/c) + (-c^-2)(c^2)cos(1/c)
=> f ' (c) = 2csin(1/c) - cos(1/c)

B) show that f is differentiable at x = 0 and show that f(0)=0
we must show that f is continuous at x = 0

note that |(x^2)sin(1/x)|<or= x^2 for all x

so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
=> lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
=> 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
so by the squeeze theorem:

lim{x-->0+}(x^2)sin(1/x) = 0
similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0

C) show that f' is not continuous at x = 0
this seems trivial to me, but i may be wrong. at x = 0, 1/x is undefined, so f ' (x) = 2xsin(1/x) - cos(1/x) is undefined and is not continuous. you can take the left and right hand limits to see this, show that they are different or something

3. Originally Posted by Jhevon

note that |(x^2)sin(1/x)|<or= x^2 for all x

so we have: -x^2 <or= (x^2)sin(1/x) <or= x^2
=> lim{x-->0+} -x^2 <or= lim{x-->0+}(x^2)sin(1/x) <or= lim{x-->0+}x^2
=> 0 <or= lim{x-->0+}(x^2)sin(1/x) <or= 0
so by the squeeze theorem:

lim{x-->0+}(x^2)sin(1/x) = 0
similarly, we can show, lim{x-->0-}(x^2)sin(1/x) = 0

thus f is continuos at x = 0, so by the definition of continuity we have f(0) = lim{x-->0}(x^2)sin(1/x) = 0
But this shows that f is continous at 0 not differenciable at 0

4. Originally Posted by ThePerfectHacker
But this shows that f is continous at 0 not differenciable at 0
there are three things that can cause a function f to be non-differentiable, is there not?

if:

(1) f is undefined
(2) f is discontinuous
(3) f has a sharp turn

case 3 doesn't happen, and case (1) doesn't happen because of the way the function is defined (by definition, f(0) = 0, so it's not undefined there as the problem is phrased), so i thought i could just take care of case (3) and be done with it, which is what i did

and i forgot to say "this is my 13th post!"
darn it, i missed it again!

5. Originally Posted by learn18
let f(X) = x^2 sin(1/x) for x is not equal to 0 and f(0)= 0

B) show that f is differentiable at x = 0 and show that f(0)=0
First f(x) is defined on an open interval containing 0.
Thus we can consider the derivative limit:

lim(x-->0) (f(x) -f(0))/(x-0) = lim(x-->0) f(x)/x

lim(x-->0) x^2 sin(1/x)/x = lim (x-->0) x*sin(1/x)

The limit exists because:

-1<=sin(1/x) <= 1 then

-x<= x*sin(1/x) <= x for x>0
x<= x*sin (1/x) <= -x for x<0
In both cases the limit as x-->0 is 0 by Squeeze Theorem.

Thus, f(x) is differenciable at 0, and furthermore the derivative is 0.