Question

lim x->infinity
(sinx+cosx)/(x^2)=?

Answer 1

\[\LARGE \frac{(1-\frac{x^2}{2}+\frac{x^4}{24})+(x-\frac{x^3}{6}+\frac{x^5}{120})}{x^2}\]
Tried using series..I got 1/2 i guess.

Answer 2

it's Easy, i think this question is repeated by you.

Answer 3

try to remove x from denomiator

Answer 4

then, replace x with 0

Answer 5

i get -1/2

Answer 6

Good

Answer 7

wrong

Answer 8

Summons @ParthKohli

Answer 9

Summon successful.

Answer 10

Summons @terenzreignz

Answer 11

Summon successful.

Answer 12

Just for that, I'm fanning you. Now, stand by... while I assess the situation.

Answer 13

I await sire.

Answer 14

Okay, I have a verdict, but I think I'll wait for ParthKohli, he seems to be putting quite a bit of effort into this :D

Answer 15

Is it so?
@ParthKohli :O

Answer 16

Why is -1/2 wrong though ?

Answer 17

You'll see. The limit is zero, by the way :)

Answer 18

hmm..

Answer 19

Hey, guys I don't know how to do this one :-)

Answer 20

LOL.

Answer 21

JK hahahaha! :-P

Answer 22

Haha, @ParthKohli
L'Hopital doesn't apply, the numerator doesn't go to infinity :P

Answer 23

Exactly.

Answer 24

*close enough*

Answer 25

My turn?

Answer 26

Nevermind.

Answer 27

But why is series going wrong?!

Answer 28

I'll just keep typing and testing, don't expect much from me in the next few minutes.

Answer 29

I have no idea, but it takes a clever bit of analysis, to realise...
\[\huge -\frac{2}{x^2}<\frac{\sin x+\cos x}{x^2}<\frac2{x^2}\]And SQUEEZE it :P
Both -2/x^2 and 2/x^2 go to zero :P

Answer 30

\[\LARGE \cancel x^2\frac{(\frac{1}{x^2}-\frac{1}{2}+\frac{x^2}{24})+(\frac{1}{x^2}-\frac{x}{6}+\frac{x^3}{120})}{\cancel x^2}\]
other terms with 1/x goes to 0..we are left with -1/2!!

Answer 31

http://www.wolframalpha.com/input/?i=limit+%28sin+x%2Bcos+x%29%2Fx%5E2+as+x-%3Einfinity

Answer 32

\[\dfrac{\sin(x)}{x^2} + \dfrac{\cos(x)}{x^2} = \dfrac{1}{x} \left(\dfrac{\sin(x)}{x} + \dfrac{\cos(x)}{x}\right)\]Yoohoo, yoohoo.

Answer 33

^^

Answer 34

I just used Squeeze. Who needs L'hopital :P

Answer 35

But why is series wrong..somenoe?!

Answer 36

Better consult an expert with series... I'm no such expert :D

Answer 37

hmm..okay I liked @ParthKohli 's solution :P seems more legit :D

Answer 38

No matter :D
As long as you agree that it's zero.

Answer 39

\[\LARGE \frac{xtan2x-2xtanx}{(1-\cos2x)^2}\]
x->0
anyone? :P

Answer 40

Okay, L'Hopital looks good here.

Answer 41

I have a doubt.

Answer 42

We won't use quotient rule here right?

Answer 43

The hell with the product rule. >_>

Answer 44

Just differentiate them seperately?

Answer 45

Nope, just differentiate the numerator and the denominator.

Answer 46

@DLS You cancelled the \(x^2\)s? Where did the \(x^2\) in the numerator come from? o_O

Answer 47

Factored out an x^2 from the series.

Answer 48

^^

Answer 49

Ah, thanks

Answer 50

so i get 1/x terms which tend to 0.

Answer 51

OH GUYS SORRY!! SERIES CANT BE USED HERE >.<

Answer 52

Series are only applicable for x<1
Foolish me.!!

Answer 53

Only two of them, you still get terms with x^n in the numerators, some positive, some negative.

@DLS Why complicate limits with series? :P

Answer 54

Stay within your LIMITS, or you'd get a SERIES of disasters!

Answer 55

haha

Answer 56

@DLS, try this http://www.freemathhelp.com/forum/archive/index.php/t-45901.html?s=81349e7c156e89af0e48100743b7ef6d

Answer 57

yaayyy! free maths help!!

Answer 58

What's OS for, then? :/

Answer 59

OS is for free math cheating!

Answer 60

Way to promote the site, ambassasdor ^
:D

Answer 61

^_^

Answer 62

\[\LARGE \frac{\tan2x+x^2\sec2x-2tanx-2xsec^2x}{2(1-\cos2x) \times?}\]
not sure how to do the denominator :/ not sure about numerator either :P

Answer 63

http://wolframalpha.com

Answer 64

is this a limits again?

Answer 65

16sin^3xcosx how :/

Answer 66

\[\huge \frac{d}{dx}(1-\cos 2x)=2\sin2x\]

Answer 67

^ Chain Rule :')

Answer 68

The answer would be 1/2 when x-->0

Answer 69

but its not so

Answer 70

:-O RnR?

Answer 71

I was bored man. @ParthKohli

Answer 72

I'm going to go for a bit again... be back later, maybe ^.^
---------------------------------------------------
Terence out

Answer 73

:(

Answer 74

):

Answer 75

How do i get the damn answer?

Answer 76

Chain Rule aata hai?

Answer 77

haan

Answer 78

\[\LARGE \frac{x^2secx-2xsec^2x}{2(1-\cos2x) \times (2\sin2x) \times (1-\tan^2x)}\]
im left with this pellet :P

Answer 79

:-O

Answer 80

??

Answer 81

i changed tan2x to its formula,so i can eliminate it

Answer 82

Differentiate mat karo..
tan2x ko kholo=(2tanx/1-tan^2(x))
2xtanx ko bahar nikal lo.
Bach jayega [2xtanx(tan^2(x)]/[4sin^4x(1-tan^2(x)]
Isko cos aur sin me convert karo..
1/2[xtanx]/sin^2x*(cos^2x-sin^2x).
Isko solve kro 1/2 aayega.

Answer 83

galaat hai to kyun padu be :/

Answer 84

Bhaad me jaa.

Answer 85

:)