Question

continuityyyyyyyyyyyyyyyyyyyy

Answer 1

First of all for solving this question do you know the conception of : firstly function, then limit then continuity.

And i garantee you will then on your own can solve such questions :)

Answer 2

i know
but there is mod x
so we have to take +x and -x

Answer 3

we have to check somethinks here
first the limit by the right tha limit is zero
then the limit by the left tha limit is zero too
and the value of the function en x=0 is equal to the limit by the left and the right so i would say that function is contnuos

Answer 4

Uhh... i think it's \[\Large f(x) = |x| \cos \frac{ 1 }{x }, x \neq 0\]\[\Large f(x) =0, x=0\]

Check limits on both sides for the first function, as it needs to approach zero for continuity.

Answer 5

\[\Large f(x) = |x| \cos \frac{ 1 }{x }, x \neq 0\]
the cos(1/x) approaches 1 as x approaches 0. |x| approaches zero as x approaches zero. (applies to either side)

Answer 6

The thing approaching 0 can not be in the bottom of a fraction because it will not allow a simple limit evaluation. You end up with a \(\pm\infty\) problem, which means the limit DNE. So you need to get it out of the bottom or to where it does not matter because it is \(\pm\) some other value, and then as it gos to 0 it does not matter.

Answer 7

as x-> 0
cos (1/x) can take any value from 0 to 1
but |x| ->0
so the entire limit = 0

Answer 8

okay i got it
its continuous

Answer 9

yes.

Answer 10

@agent0smith
thank u

Answer 11

and @hartnn and @e.mccormick
also thank u

Answer 12

@e.mccormick exactly what @hartnn said. The cos(1/x) will be between -1 and 1 as x=>0, while |x| approaches zero, so the function approaches zero.

Answer 13

oh, yeah -1 and 1 :)

Answer 14

You know, I made that 0 to 1 mistake earlier today myself. LOL.

Answer 15

haha, well up there I mistakenly said "the cos(1/x) approaches 1 as x approaches 0" and I don't even remember why. I think I assumed (1/x) approaches 0 as x approaches zero, so cos(1/x) approaches 1... wtf O_o