Anonymous

Determine whether f'(0) exists

f(x) = {

x*sin(1/x) if x not = 0

0 if x = 0

The book answer said not exists. but i just wounder if by squeeze theorem it is exists. may be i confused something. Can any one explain.

Thank You.

f(x) = {

x(squ.)*sin(1/x) if x not = 0

0 if x = 0

what 's different ? Help ?

Update:

Thanks nelsonywm2000

Is that meant : f(x)= lim x->0 x*sin(1/x) exists.

but f'(x) not exists.

Rating

f(x) = {

x*sin(1/x) if x not = 0

0 if x = 0

f'(x) = lim[h->0] [f(x + h) - f(x)]/h

f'(0) = lim[h->0] [f(h) - f(0)] / h

f'(0) = lim[h->0] [h*sin(1/h) - 0] / h

f'(0) = lim[h->0] [sin(1/h)]

which cannot be determined, so f'(0) does not exist

f(x) = {

x(squ.)*sin(1/x) if x not = 0

0 if x = 0

f'(x) = lim[h->0] [f(x + h) - f(x)]/h

f'(0) = lim[h->0] [f(h) - f(0)] / h

f'(0) = lim[h->0] [h^2*sin(1/h) - 0] / h

f'(0) = lim[h->0] [hsin(1/h)]

Since -1 <= sin(1/h) <= 1,

lim[h->0] [-h] <= lim[h->0] [hsin(1/h)] <= lim[h->0] [h]

0 <= lim[h->0] [hsin(1/h)] <= 0

lim[h->0] [hsin(1/h)] = 0

So f'(0) exists and equals 0