Question

Find the value of the Limit :

\[\color{blue}{\lim_{x \rightarrow \infty}( \frac{x + \sin(x)}{x + \cos(x)})}\]

Answer 1

just try to apply L'hospital Rule

Answer 2

Okay:

After applying L' Hospital Rule for first time, I am getting like this:

\[\lim_{x \rightarrow \infty} (\frac{1 + \cos(x)}{1 - \sin(x)})\]

Answer 3

\[\lim_{x \rightarrow \infty} \frac{ 1+\cos x }{ 1-\sin x }\]

Answer 4

is this is write?

Answer 5

Yes it is...

Answer 6

then rationalize it,i think you can get the answer

Answer 7

After taking L'Hospital you are saying to apply Rationalization ??? @dirtymind

Answer 8

I think its 1..we have to use expansions of sinx and cosx

Answer 9

Or you are suggesting other method by using Rationalization???

Answer 10

even i say rationlize it

Answer 11

Can you explain properly @AbhimanyuPudi ..

Answer 12

by multiplying and dividing by 1-sin x

Answer 13

yes rationalize it by 1+sinx

Answer 14

1 + sin(x) @uzumakhi

Answer 15

yes apply it

Answer 16

Let us try as you are saying ..

Answer 17

Substitute these formulae:
sinx = x - x^3/3! + x^5/51....
cosx = 1 - x^2/2! + x^4/4!...

Answer 18

thanks @waterineyes

Answer 19

why wont you use sandwich ?

Answer 20

\[\frac{(1 + \cos(x)) \cdot (1 + \sin(x))}{\cos^2(x)} \implies \frac{1 + \sin(x) + \cos(x) + \sin(x) \cdot \cos(x)}{\cos^2(x)}\]

Answer 21

I don't know the answer properly..

Answer 22

you can bound it and use sandwich

\[\frac{ x-1 }{ x+1 } < .. < \frac{ x+1 }{ x-1 }\]

Answer 23

is there anything wrong here?

Answer 24

What is Sandwich?

Ca you explain it I am hearing this word in Limits for first time...

Answer 25

now take sin^2x at denominator of each function and make 1/infinity=0

Answer 26

first of all .. using this sandwich or squeeze gives 1 as the answer.

Answer 27

http://en.wikipedia.org/wiki/Squeeze_theorem

Answer 28

How can I take that @dirtymind

Answer 29

may i introduce a simpler approach ?

Answer 30

if you have limf(x) and you can bound it by g(x) and h(x)
g(x) <= f(x) <= h(x)

if limg(x) = limh(x) = L
then
limf(x) = L

Answer 31

Always @hartnn

Answer 32

put x=1/y
when x->infinity, y->0

Answer 33

so as you see lim[ (x-1) / (x+1) ] = lim [ (x+1)/ (x-1) ] = 1
so your limit is 1 .. done

Answer 34

I am getting your point @Coolsector ..

Answer 35

i also liked the sandwitch approach by coolsector

Answer 36

Yeah I got it @Coolsector

Answer 37

now it looks like that i talked to myself:P

Answer 38

Thanks for your precious help...

Answer 39

yw

Answer 40

u could also get it wothout sandwitch

Answer 41

By using your idea @hartnn ??

Answer 42

u know what is lim x->0 x sin(1/x) = ??

Answer 43

0...

Answer 44

0

Answer 45

Yeah I said the same..

Answer 46

http://www.wolframalpha.com/input/?i=the+limit+of+%28x%2Bsin%28x%29%29%2F%28x%2Bcos%28x%29%29+as+x+goes+to+infinity

Answer 47

sin(1/x) always lies between -1 and 1 and so as x approaches zero, 0<=|x sin(1/x)| <= |x| but since x->0, |x sin(1/x)| must approach zero also.
similarly for cos

Answer 48

I liked all the three discussed approaches to this problem...I wish i could give 3 medals to @hartnn , @Coolsector , @AbhimanyuPudi

Answer 49

Then what??

I am not getting properly... @hartnn

Answer 50

ok, u put y=1/x in original limit , what u get ?

Answer 51

Yes you can give 3 medals to all 3 but firstly I must understand the solutions..

Answer 52

\[\frac{\frac{1}{y} + \sin(\frac{1}{y})}{\frac{1}{y} + \cos(\frac{1}{y})}\]

Answer 53

and y->0

Answer 54

And limit y tends to 0 too..

Answer 55

so u have now (1+ysin(1/y)) in numerator , right ?

Answer 56

\[\frac{1 + y \sin(\frac{1}{y})}{1 + y \cos(\frac{1}{y})}\]

Answer 57

There limit is going to 0 so 1/1 is 1..

Is this what you are saying @hartnn

Answer 58

Or I said it in a hurry ??

Answer 59

ok, now since sin fluctuates between 1 and -1 only, but y sin (...) will be very near to 0, because y is very near to 0
so, lim y->0 ysin(1/y) = 0

Answer 60

similarly for denominator

Answer 61

I liked it...but don't we consider sin(1/0)...its not defined, right?

Answer 62

Yeah I got your approach too @hartnn

Thanks for this help also...

Answer 63

But we are multiplying it with a number nearer to 0...

Answer 64

sin(1/0) is not defined
but here y is NOT 0
its very very near to 0

Answer 65

Excellent!!! got it.

Answer 66

whoa! 8 medals given in this question!!

Answer 67

Sorry not 8..

You have mistaken in a counting..

Answer 68

u guys deserve more, very good explanations xD

Answer 69

:D

Answer 70

Total 9...

Answer 71

yeah, 1 was given after i counted :P

Answer 72

It is more like a Tutorial @hartnn

Answer 73

Hahaha...yea 9..this is first time i've seen these many medals given!!!

Answer 74

yo! 10 now.
thnx cool.
yes.

Answer 75

lol yes

Answer 76

tutorials should be like this, engaging way

Answer 77

Superb discussion!!