Question

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Answer 1

So the function is defined at x=0 but in the neighbourhood around x=0, f(x) is undefined.

Answer 2

yes...u r right

Answer 3

then use the definition of continuity..
limit does not exist so not continuous

Answer 4

does it make sense?

Answer 5

So should I work out the limit from the left and right and say that they are not equal, so the limit does not exist at x=0 so the function is not continuous at x=0

Answer 6

Or would a proof by contradiction using the delta-epsilon proof of continuity work better?

Answer 7

1st left hand or right hand limit donot exist...so there is no question of equality of lh or rh limit

Answer 8

both left hand and right hand limits are indeterminate..

Answer 9

u can use delta and epsilon proof

Answer 10

For all epsilon > 0, there exists a delta > 0 such that \[|x-0| < \delta => |f(x) - f(0)| < \epsilon \]

Answer 11

0 < I 0 - x I < delta then I f(0) - f(x) I < epsilon

Answer 12

\[|x| < \delta => |f(x)| < \epsilon \]

Answer 13

yes...

Answer 14

So if I fix epsilon to be, say (1/2), the value of x will be within delta of 0 if sin(1/x) is less than (1/2), but sin(1/x) oscillates between 1 and -1, so f(x) has values that are greater than (1/2), which is a contradiction?

Answer 15

choose any epsilon...but -1 <= f(x) >= 1

Answer 16

yes

Answer 17

Awesome!

Answer 18

Thank you so much!

Answer 19

:)

Answer 20

Might have to pick your brain in another question too!