Question

limit as x approaches infinity of 1-cosx/x^2 using sandwich theorem

Answer 1

i just have to say sandwich is a funny name.

Answer 2

that was helpful..

Answer 3

sorry...

Answer 4

since \(-1\leq\cos(x)\leq 1\) we know \(-2\leq 1-\cos(x)\leq 0\)

Answer 5

ok my inequality stinks, let me try again

Answer 6

\[0\leq 1-\cos(x)\leq 2\] that looks better

Answer 7

where'd you get the second inequality?

Answer 8

which means can "sandwich"
\[0\leq \frac{1-\cos(x)}{x^2}\leq \frac{2}{x^2}\]

Answer 9

mmmm sandwich
oh first inequality is wrong

Answer 10

\[-1\leq \cos(x)\leq 1\] is right
and so \(1-\cos(x)\) is smallest when \(\cos(x)=1\) in which case you get 0, and largest when \(\cos(x)=-1\) in which case you get 2

Answer 11

pretty clear that as \(x\to \infty\) we know \(\frac{2}{x^2}\to 0\) so you have a zero on the left and a zero on the right and that is your sandwich

Answer 12

wait I don't understand how cos(x) = 1 you get 0? and such?

Answer 13

ok lets go slow

Answer 14

the largest cosine can be is 1, so the smallest -cosine can be is -1 and so the smallest 1-cosine can be is 0

Answer 15

this is really messing me up, if this is only the beginning of calculus, i'm scared

Answer 16

i think you are fretting for very little reason. calc can be difficult, but this part is just reasoning, not computation
is it clear that the smallest 1 - cosine can be is 0 ?

Answer 17

it is because cosine is stuck between minus one and one, and therefore so is - cosine

Answer 18

it's not making sense.. 1 - (-1) is 2....

Answer 19

yes exactly, that is the very largest \(1-\cos(x)\) can be
the biggest it can be is if cosine is minus one, and in that case you get 2
that is the maximum it can be

Answer 20

in other words \(1-\cos(x)\) cannot bet greater than 2
you have just shown it

Answer 21

so cos(x) can't be greater than 1, but that doesn't mean 1-cos(x) can't be greater than 1

Answer 22

yes you are right. it doesn't mean \(1-\cos(x)\) cannot be greater than one, it means \(1-\cos(x)\) cannot be greater than 2 as you said above

Answer 23

okay so \[1-\cos(x) \le 2\]

Answer 24

yes for sure. it is more or less clear right?

Answer 25

alright, continue, it's catching on

Answer 26

perhaps the 1 at the beginning is confusing
suppose instead we had \(10-\cos(x)\) what is the biggest that could be? answer is 11, because of the bound on the cosine
what is the smallest? answer is 9 for the same reason

Answer 27

I didn't know there was a sandwich theorem! Thanks @satellite73 and @jamroz

Answer 28

in any case we see that the very largest \(1-\cos(x)\) can be is 2
what is the smallest it could be? answer is 0. it would be zero if \(\cos(x)=1\) you would get \(1-1=0\) and it cannot be any less than 0, because cosine cannot be any larger than 1

Answer 29

okay that makes sense, understanding that, what do I do now?

Answer 30

not that hold on, new paragraph to read

Answer 31

ok let me know when we continue

Answer 32

okay got it, now what?

Answer 33

thank you so much for your patience by the way, really appreciated

Answer 34

no problem

Answer 35

ok now we almost have our sandwich. \(1-\cos(x)\) is stuck between 0 and 2, so we write the inequality
\[0\leq 1-\cos(x)\leq 2\]

Answer 36

i have that down, now what?

Answer 37

then dividing by \(x^2\) gives
\[0\leq \frac{1-\cos(x)}{x^2}\leq \frac{2}{x^2}\]

Answer 38

so far so good?

Answer 39

understood, proceed

Answer 40

then we take the limit as \(x\to \infty\)

Answer 41

on the left we get 0 because there isn't even an \(x\) in it, it is just always 0

Answer 42

and what about
\[\lim_{x\to \infty}\frac{2}{x^2}\]? well that has to be 0 as well, because \(x^2\) is getting larger and larger, but the numerator is stuck at 2

Answer 43

I don't understand why it has to be 0?

Answer 44

oops i meant
\[\lim_{x\to \infty}\frac{2}{x^2}=0\]

Answer 45

\[\lim_{x \rightarrow \infty} 2\div x^{2}\]

Answer 46

why does that have to be 0?

Answer 47

couldn't it still be 1 and work?

Answer 48

x is getting larger and larger. think of a number with a huge denominator and a 2 in the numerator
that number is going to be very close to 0
for example if \(x=10\) then \(x^2=100\) and \(\frac{2}{100}=.02\) and that is only if \(x=10\) not a very large number
now suppose \(x=1000\) then \(x^2=1,000,000\) and \(\frac{2}{x^2}=0.000002\) getting very close to 0

Answer 49

but if x = 1 then x^2 = 1 and it'd be 2/1 = 2?

Answer 50

brb, away from the keyboard for a minute

Answer 51

you can make it as close to zero as you like, by making x larger and larger
the bigger x gets, the smaller \(\frac{2}{x^2}\) gets

Answer 52

oh yes, at 1 you get two but you are taking the limit as x goes to infinity, not 1
x is getting bigger and bigger, not closer and closer to 1

Answer 53

true, true. you make your point. so that's how I validate it's going to 0 using sandwich?

Answer 54

i'm trying one for the lim x as it approaches -infinity for sinx/x
and i'm using the sandwich kind of.. is this the right path?
\[-1 \le sinx \le 1\]
\[\frac{ -1 }{ x } \le \frac{ sinx }{ x } \le \frac{ 1 }{ x }\]

Answer 55

yes

Answer 56

then where would I go from there? it's catching on thank you so so much

Answer 57

to answer your question above, since the limit on the left is 0 and the limit on the right is 0, the limit in the middle must be 0 as well

Answer 58

\[0\leq \lim_{x\to \infty}\frac{1-\cos(x)}{x^2}\leq 0\] indicates that the middle one must be zero as well

Answer 59

that makes complete sense! :D

Answer 60

\[\frac{ -1 }{ x } \le \frac{ \sin(x) }{ x } \le \frac{ 1 }{ x }\] just like the previous one
take the limit as \(x\to \infty\) on the right and the left
you will get 0 in both places

Answer 61

you will have to take my word for it that soon this will all seem completely obvious
since sine is stuck between -1 and 1, as the \(x\) in the denominator gets larger and larger the whole thing must go to zero

Answer 62

because with -1/x when x gets larger it gets closer and closer to 0 as it approaches -infinity right?

Answer 63

yes for sure

Answer 64

the minus one up top is a red herring. it could be 57,208 and it would still go to zero as the denominator goes to minus infinity

Answer 65

but isn't sinx/x = 1?

Answer 66

that is the limit as x goes to zero, not as x goes to infinity

Answer 67

\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\] but
\[\lim_{x\to \infty}\frac{\sin(x)}{x}=0\]