Question

similar to last: Prove that h(x) is not differentiable using the definition and coming to a contradiction.
\[h(x)=\{ 0~if~x<0;~~x~if~x\ge 0\}\]

Answer 1

so, here is what I got:
let \(x_n=\frac{1}{n}\) and \(y_n=\frac{-1}{n}\) then \(\lim_{n\rightarrow \infty} x_n ,y_n=0\). Then by the sequential definition of a limit, if the limit exists, then \(\lim_{n\rightarrow \infty} h(x_n)=lim_{n\rightarrow \infty} h(y_n)\)

Answer 2

then i want to apply that to the limit def of derivative

Answer 3

but I don't know how

Answer 4

@eliassaab

Answer 5

so the limit definition is this: \[\lim_{x\rightarrow c} \frac{f(x)-f(c)}{x-c}=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)\]

Answer 6

@ganeshie8 I know this is out of your area, but any insight?

Answer 7

@Loser66

Answer 8

do you get \(h(y_n) = 0\) and \(h(x_n) = -\frac{1}{n}\) ?

Answer 9

?

Answer 10

both limits are 0, so i dont see how exactly your strategy works in this proof

Answer 11

well, I'm not allowed to use a left or right limit, and the TA used them so I have no clue how I am to prove it, but it is derivative, not continuity. The left limit will be 0 and the right 1

Answer 12

you want to show \(\lim \frac{h(x_n)-h(0)}{x_n-0} \ne \lim \frac{h(y_n)-h(0)}{y_n-0}\) right ?

Answer 13

yea

Answer 14

alright, i was confusing continuous with differentiability earlier
let me think a bit freshly..

Answer 15

so the hint I was just given by a classmate is to use |x_n-y_n|<delta where x_n=n and y_n=n-1/n

Answer 16

It has been found!

Answer 17

so the missing step waas to evalualte the limits for lim xgoes to inftyf(xn)-f(0)/x_n

Answer 18

here is my attempt
Claim: function is not differentiable at x = 0
proof
By contradiction
Assume that the derivative does exist at x =0.

Therefore by definition of derivative,
limit [f ( 0 + h) - f(h) ]/ h exists as h->0

but for a limit to exist at a point, the right limit and left limit must match (equal the same number).

lim {h->0-} [f(0+ h) - f(0 )] / h = lim {h-> 0+} [f(0+ h) - f(0)] / h



but
lim {h->0-} [f(0+ h) - f(0 )] / h = (0- 0 ) / h = 0

and

lim {h-> 0+} [f(0+ h) - f(0)] / h = (h - 0 / h) = 1

therefore we have a contradiction , 0 = 1

Answer 19

are you still using ur old definitions of xn and yn ? can you show ur proof using sequence definition of limit ?

Answer 20

@FibonacciChick666

Answer 21

yea,

Answer 22

yes please

Answer 23

Okay so i can disregard this hint ?

`so the hint I was just given by a classmate is to use |x_n-y_n|<delta where x_n=n and y_n=n-1/n `

Answer 24

maybe you could substitute 1/n , and -1/n for h in the derivative definition, since they both go to zero