Question

The function f(x) is continuous at 0. Show that f is differentiable at 0, and find f prime at 0.
Hint: Look at the Newton quotient for f at 0 and use L’Hospital’s rule.

Answer 1

f(x) = (sin x)/x when x is not = 0
= 1 when x = 0

Answer 2

where are you stuck?

Answer 3

not sure where to start

Answer 4

look at

\[\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\]

Answer 5

use newtons quotient

Answer 6

so after applying HR i get the limit as x->0 [xcos(x)-sin(x)]/x^2

Answer 7

u used lhopitals?
you sure

\[\frac{f'(x)}{g'(x)}\]

Answer 8

nvm i got it

Answer 9

i got 0

Answer 10

f'(x)= cos(x)/1 at 0 = 1

Answer 11

0 is correct

Answer 12

so that shows it's differentiable at 0?

Answer 13

yes...if the limit did not exist then it would not be differentiable....the limit did exist so it is differentiable

Answer 14

i'm so confused here are you guys talking about newtons quotient?

Answer 15

ok ty

Answer 16

so now what is f prime at 0?

Answer 17

0

Answer 18

how?

Answer 19

it is the limit you got

Answer 20

oh i see thanks