- #1

StephenPrivitera

- 363

- 0

f(x)={g(x)sin(1/x), x not =0

=0, x=0

where g(0)=g'(0)=0.

f'(0)=g'(0)sin(1/0) + g(0)dsin(1/x)/dx

=0sin1/0+0dsin(1/x)/dx=0?

applying the limit definition, I get

f'(0)=g'(0)lim sin(1/h) where h-->0

is this zero?

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- Thread starter StephenPrivitera
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- #1

StephenPrivitera

- 363

- 0

f(x)={g(x)sin(1/x), x not =0

=0, x=0

where g(0)=g'(0)=0.

f'(0)=g'(0)sin(1/0) + g(0)dsin(1/x)/dx

=0sin1/0+0dsin(1/x)/dx=0?

applying the limit definition, I get

f'(0)=g'(0)lim sin(1/h) where h-->0

is this zero?

- #2

StephenPrivitera

- 363

- 0

- #3

NateTG

Science Advisor

Homework Helper

- 2,452

- 7

Product rule:

f(x)g(x)=f'(x)g(x)+f(x)g'(x)

You need derivatives of both.

Now we have

g(x)sin(1/x)

the derivative of sin(1/x) is

x

and not defined at zero (no limit at zero either)

you'd need g(x) to grow at better than x

- #4

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