Question

ITS FROM DIFFERENTIATION CHAPTER:-
SHOW THAT f(x)={xsin1/x , x is not equal to 0
{0 , x is equal to 0 is continuous but not differentiable at x equal to 0.

Answer 1

If i remember this is one of those where you find the limits as it goes to 0 to show that it's not differentiable and then show that any other number the limit goes to a specific number

Answer 2

CAN U WORK OUT N EXPLAIN BRIEFLY?

Answer 3

i'd have to look it up. it's been a while however i remember doing this for calc 3

Answer 4

OK.............. DO IT. TAKE UR TIME

Answer 5

can you use lehospital rule?

Answer 6

can you make sure that the top equation is xsin(1)/x?

Answer 7

for one find the the limit when x-> 0

Answer 8

use L'opital's Rule. Take the derivative of the numerator and the derivative of the denominator,, then do direct substitution of 0.

Answer 9

show it

Answer 10

The equation is diferentiable since it is composed of continuos functions in R\{0}

Answer 11

please everyone show the work out

Answer 12

and
\[\frac{d(xsin(\frac{1}{x})}{dx}=\sin(\frac{1}{x})-\frac{\cos(\frac{1}{x})}{x}\]

Answer 13

1) definition of limit x->c f(x) = L

for any ε>0, small as you wish, there is a δ>0 such that for every x that has |x-c|<δ you have that |f(x)-L|<ε

therefore you also have that |k*f(x)-k*L|<k*ε

just set the initial ε at 1/k of your desired ε and you have the proof


lim (x->0) f(x) = 0 (sin(1/x) is limited between 1 and -1 while x goes to 0),

lim (x->0) f(x) = f(0). 2a) proved

f'(0)=lim (h->0) [f(h)-f(0)]/h = lim (h->0) f(h)/h = lim (h->0) sin(1/h) and the limit does not exist, the function oscillates infinitely between 1 and -1 as h goes to 0. alternatively put, definte t=1/h, lim (h->0) sin (1/h) = lim t-> infinity sin(t) and you can easily see that it does not exist.

3) the function is x for x>=0 and -x for x<0. The right derivative is +1 and the left is -1. They are different, therefore it is not differentiable

Answer 14

thanx allll