Question

Another Limit question

Answer 1

\[\large \bf \lim_{x \rightarrow \infty} \cos(\sqrt{x+1})-\cos(\sqrt{x})\]

Answer 2

infinity - infinity is indeterminate

Answer 3

yup !
then we can convert it into infinity/infinity form !

Answer 4

well I know the following:
\[-1 \le \cos(\sqrt{x+1})\le 1 \\ -1 \le \cos(\sqrt{x}) \le 1 \\ \text{ so it should be that } \\-2 \le \cos(\sqrt{x+1})-\cos(\sqrt{x}) \le 2 \]
well it isn't exactly infinity-infinity
both functions oscillate between -1 and 1

Answer 5

how did you get [-2,2] ?

Answer 6

played with numbers in the inequalities above
I was like the max of cos(sqrt(x+1)) is 1
and the max of cos(sqrt(x)) is 1
so the max of cos(sqrt(x+1))+cos(sqrt(x)) could be (1+1)=2

and then the same thing with the min values for each

Answer 7

but you subtract 2 equations ! and you actually add them

Answer 8

or I also I mean to put a minus there
cos(sqrt(x+1))-cos(sqrt(x) could be (1+1)=2
you know since -cos(sqrt(x)) could be 1

Answer 9

-cos(sqrt(x)) is actually 1 when x=pi^2

Answer 10

or that is one value anyways

Answer 11

ok. so how can we solve limit ?

Answer 12

OMG NOOOO IT'S NOT INDETERMINATE

Answer 13

omfg omfg omfg omfg SQUEEZE THEORUM

Answer 14

NO GUYS WTF ARE YOU GUYS DOING

Answer 15

SQUEEZE THEORUM

Answer 16

i don't know about this theorem?

Answer 17

i am a LEARNER !

Answer 18

Oh Okay, good I can teach it to you

Answer 19

\[\infty-\infty\]? seems unlikely

Answer 20

Misty I got this haha, let me explain squeeze to her

Answer 21

So basically what's the range for cosx

Answer 22

her ?
i am a boy

Answer 23

it doesn't have a squeeze to it darling q

Answer 24

lol

Answer 25

Yes it does....

Answer 26

Any trig function taken to limit as x-->infinity has to be squeezed

Answer 27

try this substitution
x = tan^2(y)

Answer 28

lol sure, i like being squeezed too, but that doesn't mean i have to be squeezed

Answer 29

Mayank.. If you want to listen to what these guys are incorrectly telling you go for it. I can help you otherwise.

Answer 30

it does just what @freckles said, goes between -2 and 2
no limit just upper and lower bounds

Answer 31

can we apply cos C - cos D ?

Answer 32

only if you want to make it look uglier

Answer 33

can we rationalize it ?

Answer 34

what formula is that

Answer 35

http://www.wolframalpha.com/input/?i=limit%28-2*sin%28%28sqrt%28x%2B1%29%2Bsqrt%28x%29%29%2F2%29*sin%28%28sqrt%28x%2B1%29-sqrt%28x%29%29%2F2%29%2Cx%3Dinfty%29

weird thing is wolfram says the limit exists and is 0 if we rewrite it like that

Answer 36

I did write it correctly right?

Answer 37

cos C - cos D = 2 sin (C+D/2) sin ( D-C/2)

Answer 38

can we rationalize it guys ?

Answer 39

http://www.wolframalpha.com/input/?i=lim%28x-%3Einfinity%29+cos%28sqrt%28x%2B1%29%29+-+cos%28sqrt%28x%29%29

Answer 40

you have a mistake in your formula, it should be this
http://www.wolframalpha.com/input/?i=+cos%28sqrt%28x%2B1%29%29+-+cos%28sqrt%28x%29%29

Answer 41

where is mistake in my formula ?

Answer 42

sin(x) is an odd function

Answer 43

yup !

Answer 44

sin(-x)=-sin(x)

Answer 45

we understood !

Answer 46

sorry, that seems to work

Answer 47

I don't get why there are answers that are contradicting each other

Answer 48

I still think the limit oscillates between -2 and 2

Answer 49

we can always plug in large numbers

Answer 50

yes it converges to zero

Answer 51

numerically

Answer 52

I GOT MY ANSWER !

Answer 53

hurray !!!

Answer 54

you can rationalize that sine expression inside

Answer 55

nope !

Answer 56

can i post my solution ? ( if you wants)

Answer 57

sure

Answer 58

first apply cos C - cos D

Answer 59

ok