Question

what is the lim --> 1 of : (1-x)tan[(pi*x)/2]

Answer 1

(1-x)tan[(pi*x)/2] is defined at x=0

Answer 2

sorry as x goes to 1

Answer 3

ah, that makes it more interesting

Answer 4

so first off, we want to split this into sin and cos
notice that the sine limit can be taken immediately, then see what you have left

Answer 5

still with me here?

Answer 6

yes

Answer 7

tell me if you can read this\[\lim_{x\to1}f(x)\]^^^^ ?

Answer 8

yes

Answer 9

great, can you try writing out what you get after changing tan to sin/cos ?
we'll take it from there

Answer 10

i did this on my notebook or should i write it here

Answer 11

whatever is faster for you, attache the file or write it here

Answer 12

sorry i was setting up my cloud for this https://an2wfg.by3302.livefilestore.com/y2mGsQJUo8-QLVI2nqI95X0CH18DsJjVKPCagEUAqzvgP_sz0aqZbIFateb48UdsF5cVk9EZwqPjWFM8sLxQGDxOZSgvGQ8OBPdLJaw1Mm_XldbOzs6ll6glP46JrxMMiCc/IMAG0456.jpg?psid=1

Answer 13

I wouldn't have distributed the 1-x, but so far so good
now notice that you can take the limits of the sin terms already, which leaves you with what?

Answer 14

YES I GET 1 but on the cos i still get zero

Answer 15

slow down, just do the sin terms first

Answer 16

when the sin terms limits are taken what is left over?
do NOT take the limit of the cosine term yet, and put it all over a common denominator

Answer 17

i f i take only the limit of the sin i get 1-1=0

Answer 18

no, if you take the limits of the sin terms on the paper you gave attached, you get\[\lim_{x\to1}\frac1{\cos{\pi x\over 2}}-\frac x{\cos{\pi x\over 2}}=\lim_{x\to1}{1-x\over\cos{\pi x\over 2}}\]

Answer 19

i shouldn't have distributed to (1-x)

Answer 20

I said that earlier, but we can always work it backwards, so no worries :)
now two limits that are usually just memorized (because proof without l'hospital is geometric and somewhat painful)\[\lim_{x\to0}\frac{\sin x}x=1\]and\[\lim_{x\to1}{\cos x\over1- x}=0\]hopefully you already know these

Answer 21

the game now is to rearrange what we have left so it looks like one of those two forms, then we can use those limits to get the anwer to our problem

Answer 22

this is, admittedly, the tricky part. There may be a faster way to do it than what I did, but why don't you play with that for a bit and see if you can get anywhere. I'll help you more once you get stuck.

Answer 23

wait a second please

Answer 24

take your time

Answer 25

i know (cosx-1)/x equals zero

Answer 26

i am not sure about the what you have written cosx/(1-x)=0

Answer 27

same thing as the second one I wrote

Answer 28

either way, we are actually going to use the first one (well, at least I did)

Answer 29

even though at first glance it looks more like the second, but there is a trig identity that gets us from sin to cos....

Answer 30

ok

Answer 31

oh I'm sorry I did write that second one wrong :P\[\lim_{x\to0}{1-\cos x\over x}=0\]
but as I said, we're not going to use it

Answer 32

let me know if you want a hint

Answer 33

should i multiply with cos(pix/2) up and down and than use the Half-Angle Identities for cos(x/2)

Answer 34

not how I did it, but I can't say it won't work until I try it
sounds harder than my way though

Answer 35

ok tell me your way ,i will try this later

Answer 36

let y=x-1
play with that for some time...

Answer 37

sorry, let y=1-x

Answer 38

i got lim of y..>0 y/[cos(pi(y+1)/2)]

Answer 39

good, now distribute pi/2 in the cos argument

Answer 40

like I said though, use y=1-x

Answer 41

so should be a - sign there, my bad

Answer 42

yeah cos((piy+y)/2)

Answer 43

am i write about the part of lim y ...>0 ????

Answer 44

yes

Answer 45

but try that substitution and distribution again with y=1-x

Answer 46

lim y..>0 y/cos(pi/2 - ypi/2
)

Answer 47

nice, now do you see a trig identity on the bottom yet?

Answer 48

the Half-Angle Identity?

Answer 49

simpler
cos(x) becomes sin(x), and vice-versa, through a simple formula

Answer 50

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 51

Cofunction Formulas
?

Answer 52

yes

Answer 53

never seen that before .....so y/sin(ypi/2)

Answer 54

we multiply with pi/2 two up and down and get 2/pi :)

Answer 55

sorry I stepped away for a moment
yes, right answer :)
cofunction formulas are handy, you should look at a unit circle and convince yourself they are true. it's easy to see geometrically
again, nice job, see ya!

Answer 56

hey

Answer 57

i saw you are a moderator here

Answer 58

hey hey

Answer 59

this is my first time in this forum ,,it is very nice the way it works , getting live help but isn't this very time consuming , you are not charging anything besides the ads that flow around ???

Answer 60

No, though you can sign up to be a special member of some type for 10$/month (which I have not done). I don't work for OpenStudy, I am just a long-time user who they elected to police the place. That is, they don't pay me hehe.
OS is not a non-profit, but we also get a lot of grants from places like the Gates and Narional Science foundations

Answer 61

they didn't even use to have ads when I started 3 years ago ;(

Answer 62

i have seen a lot of website that charge a whole lot of money for help like this ,you do it for free.......even if they charge small amounts like 50 cents for simple problems like this people would be ready to pay small ammounts like that ,

Answer 63

ok thank you for your help and the info

Answer 64

Well, I've heard of such things, but those guys do the work for you mostly from what I've seen.The philosophy here is student teaching student. I'm glad you get a lot out of it, but beware, plenty of people will just spit out answers hoping for medals with no explanation. My job as a moderator is to curb that type of abuse.
Feel free to tag me with anymore questions, see ya!

Answer 65

great see you around

Answer 66

i will asksometime and sometimes try to help someone else when i have time .....