Question

Find the limits. Question below.

Answer 1

\[\lim_{x \rightarrow \infty}\sin^{-1} (\frac{ x }{ 1-2x })\]

Answer 2

try plugging
x= 1/y first

as x->infinity
y->0

so, what about x/ (1-2x) in terms of y ?

Answer 3

ok, i will try plugging in, but i dont understand what you mean x/(1-2x) in terms of y

Answer 4

put x= 1/y
in x/ (1-2x) and simplify

Answer 5

ok, why do we have to plug in 1/y? where do you get that from?

Answer 6

i get the part if x -> infinity, then y ->0. but not the part where you have to plug in 1/y

Answer 7

the fact that 1/x will be 0

you can do this keeping everything in x
and putting 1/x as 0 because x-> infinity, 1/x -> 0

taking a new variable y is just for convenience

Answer 8

so you made 1/y up?

Answer 9

yeah, it was quite obvious, right ?
instead of plugging 1/x = 0
now i will plug y=0, easier now

Answer 10

Ok, so I tried plugging in, my steps were

\[\lim_{x \rightarrow \infty}\sin^{-1} (\frac{\frac{ 1 }{ y } }{ 1-2(\frac{ 1 }{ y }) })\]
\[\lim_{x \rightarrow \infty}\sin^{-1}(\frac{ 1 }{ y })(\frac{ 1 }{ 1-\frac{ 2 }{ y } })\]
\[\lim_{x \rightarrow \infty}\sin^{-1} \frac{ 1 }{ y-2 }\]

Answer 11

and now im stuck again. lol

Answer 12

you change x->infinity
to y-> 0 too!

then just plug in y= 0 in that expression, and that would be it!

Answer 13

is it "legit" to change y ->0?

Answer 14

like literally just change it and not make any annotations?

Answer 15

since x= 1/y

when x-> infinity, y will tend to 0
give this as a reason alongwith, and its a perfectly legal step :)

Answer 16

oh, ok i worked it out on paper, (srly too long to even "equation it" on here, and i got -1/2, is that correct?

Answer 17

you mean you got sin^{-1} (-1/2)
right ?

Answer 18

yeah, i totally forgot about the inverse sin, lemme finish the problem real quick.

Answer 19

ok, i got -pi/6.

Answer 20

i too got the same, -pi/6 :)

Answer 21

now theres another problem i need help with,

\[\lim_{x \rightarrow \infty}e ^{\sin x}\]

Answer 22

does it follow the same concept or is it different?

Answer 23

naah, its different

Answer 24

so how would i start that problem

Answer 25

you can write that as e^ {lim x->infinity sin x}

Answer 26

because 'e' exponential function is continuous

Answer 27

oh, so you limits can basically pass thru everything?

Answer 28

cross out the you, didnt mean to type the 'you'

Answer 29

not everything, just through "continuous functions"

Answer 30

oh, ic ic. lemme see, as x-> infinity sin x, isnt that 1?

Answer 31

so its e^1?

Answer 32

nopes,
as x-> infinity sin x can be any value between 1 and -1

so, your limit can be any value between e and 1/e
got this ?

Answer 33

between e and 1/e, uhhhhhh hmmmmmm how u get that

Answer 34

because sin x can take any value between 1 and -1 when x is very large (tending to infinity)

so, e^ sin x will become e^(+1) to e^(-1)
that is from e to 1/e

Answer 35

OHHHHHH omg i get it lemme work it out on literal paper. hold up

Answer 36

sure, take your time :)

Answer 37

ok, so now we to eliminate an answer, how do we do that?

Answer 38

no :O
thats a legit final answer
e to 1/e

Answer 39

Really??? Becuz i check the back for answers after im done or when im just stuck to the max and the answer in the back is 1. ermmmmm

Answer 40

is it ?

lim x->infinity e^sin x =1
!!
i don't think so.....

Answer 41

im not sure, the book has answers to odd numbered question and it was an odd, i quadripled checked and it says 1 as the answer, ermm

Answer 42

e^(anything) =1
only if anything =0, right??

so, that means sin x =0 when x->infinity!
now this isn't true, we can never get a single value of sin x when x->infinity....

Answer 43

uhhhhhh ima just skip this one and ask my teacher. thanks for your help, can i come back to this post and ask another question later?

Answer 44

yes, but more preferable thing to do is close this post and ask new question in new post
maybe if i am not online to help, some1 online will surely be able to help you :)

Answer 45

ok, thanks for everything! see you around!

Answer 46

welcome ^_^