Question

Let f(x) = floor(cosx).
For what values a does lim x approaches a f(x) exists?

Answer 1

have you tried this on your own ? post your attempt.

Answer 2

well the definition of cosx is -1<cosx<1

Answer 3

I used a software and the answer is DNE..but i dont know how.

Answer 4

Trust yourself, @uri not the software :D

Answer 5

Lol, well I don't know how to solve it :P

Answer 6

I almost envy your uhh... popularity, :D

Answer 7

Hehe :)

Answer 8

Do as instructed, though, @uri and post your own attempt..

Answer 9

I did!

Answer 10

Any hints on how to start?

Answer 11

The definition of cosine is -1<cosx<1
Is this a good start?

Answer 12

-1 <= cosx <=1

Answer 13

^that. What @shubhamsrg said :)

Answer 14

yeah! forgot the equal or less than sign.

Answer 15

^that. What @terenzreignz said :)

Answer 16

this could go on forever, @ParthKohli
anyway... please tell me you know what the floor function is...

Answer 17

Yes, I do.

Answer 18

the floor function is the greatest integer function.

Answer 19

bloody tags... @uri ?

Answer 20

Pardon?

Answer 21

nothing... you beat me to it, @uri please proceed with what you were doing :)

Answer 22

Ok, anyone who knows how to solve this question able to guide me through?

Answer 23

Does \(\lim(x)\) mean \(\lim _{x \to \infty} f(x)\)? If so, I know this.

Answer 24

\[\lim_{x \rightarrow a} floor[cosx]\]

Answer 25

We will only be able to guide you when you tell us what you did, what you tried? Have you tried to apply the knowledge that you have about this topic in this question? I

Answer 26

Consider cases.

Answer 27

@mathslover: I did; please scroll up.

Answer 28

OK, so what if \(\cos(x)\) is an integer?

Answer 29

\[\cos(a)\]**

Answer 30

Is that equation only what you tried?

Answer 31

\[\huge \lim_{x \rightarrow a}\left\lfloor \cos(x)\right\rfloor\]
LOL

Answer 32

LOL terenz.

Answer 33

@mathslover Why are you laughing...

Answer 34

Why so serious, @ParthKohli

Answer 35

\[a = 0\]

Answer 36

Ok I will not laugh now! :)

Answer 37

No, I was just asking what he found so funny...

Answer 38

I refuse to not not laugh. To h*ll with your seriousness :)

Answer 39

LOL :-P

Answer 40

Ok no we shall come to the topic, uri's quest.

Answer 41

@uri Consider\[a = 0\]

Answer 42

ok so cos(o) = 1

Answer 43

yes

Answer 44

my knowldege of floor functions are rusty tho.

Answer 45

Visualise!!!

Answer 46

How?

Answer 47

Imagine the graph, I guess :)

Answer 48

@terenzreignz did it :-D

Answer 49

I just got the answer.

Answer 50

How can I imagine the floor function?

Answer 51

No I didn't :/

Answer 52

The graph of a floor function is just like "steps".

Answer 53

Draw it.

Answer 54

okk

Answer 55

forget it!

Answer 56

closing quesion..

Answer 57

but @ParthKohli was just getting to the good part :(

Answer 58

Asks difficult integrals.

Can't draw a graph.

Answer 59

lol

Answer 60

;-)