Question

Lim (x-sinx/x) sin(1/x)
When x----->0 ...

Answer 1

The division sign is under the whole term x-sinx and not just sinx

Answer 2

is it this question!

Answer 3

@faiqraees

Answer 4

Nope!...
It is sin(1/x) this term is in multiplication with first one..
And even x tends to 0

Answer 5

you're stuck only fir sin(1/x) term right?

what is the range of sine function?

Answer 6

Exactly!
(-1,1)

Answer 7

Oops...closed bracket i meant...

Answer 8

ok,
so whatever number x tends to
sin (1/x) will ALWAYS be withing that finite range.
agree?

Answer 9

within*

Answer 10

It will be oscillating at an indeed rapid pace between the values -1,1

Answer 11

it doesn't exist. and it diverges.

Answer 12

right, we'll take your words :)

what is this product now:
0*(number oscillating at an indeed rapid pace between the values -1,1) = ...?

Answer 13

Bt this is not exact 0...the term u multiplied it with

Answer 14

the 2nd term?
its a finite number right?

Answer 15

Nope...the first one

Answer 16

what?

1- sin x/x = 1-1 = 0
basics of limits :P

Answer 17

is it this!

Answer 18

Sinx/x <1

Answer 19

can you write the actual question which make sense .. put bracket to distinguish!

Answer 20

ok, let me put it this way:
\(\lim \limits_{x \to 0}\dfrac{\sin x}{x}=1\)
agree?

Answer 21

Yep...

Answer 22

Do we need to separate limits like this
Lim (f(x)*g(x)) can be written as
Lim (f(x)) *lim(g(x))

Answer 23

exactly

Answer 24

Ok fine then...it"s done...

Answer 25

Thanks...

Answer 26

you need to use the squeeze theorem

Answer 27

Thanks sir...! Bt it's an mcq type question