Question

find limit of... limt x-->1 [2x](x-1).....and points of continuity of f(x)=x-[x], for all x belongs to real no..and please what does it mean by [x]?

Answer 1

i bet it is greatest integer function

Answer 2

limit should be 2

Answer 3

nevermind...errr

Answer 4

\[[x]\] is the greatest integer less than or equal to x. so you have
\[[\pi]=3\] and
\[[-\pi]=-4\]

Answer 5

here is a nice picture of
\[f(x)=x-[x]\]
http://www.wolframalpha.com/input/?i=x-+greatest+integer%28x%29

Answer 6

discontinuous at the integers

Answer 7

i need a sol problem and i have ashed 2 different question along with the meaning of [x]..please help

Answer 8

ok here is a picture of
\[[x]\] by itself
http://www.wolframalpha.com/input/?i=greatest+integer+%28x%29

Answer 9

it is called a "step function" because the graph looks like steps.

Answer 10

it means take the biggest integer less than or equal to x. for example,
\[[2.3]=2\]
\[[7.999]=7\]
\[[-2.4]=-3\] in each case the output is the largest integer less than or equal to the input

Answer 11

it is discontinuous at the integers because it jumps there. it is constant on the interval
\[[n, n+1)\] where n is an integer.

Answer 12

so for example on the interval
\[[3,4)\] it is always 3

Answer 13

let discuss first question...lim x ->1 [2x](x-1)..so what i have to do...i will have to put 1 or some intiger greater than 1

Answer 14

this is
\[[2x](x-1)\] right?

Answer 15

you have to take the limit from the right and from the left and see if you get the same thing or not.

Answer 16

yes...i have to calculate just limit for that function,,no need of right or left

Answer 17

if 1<x then
\[[2x]=2\]
so you get
\[2(x-1)\] which goes to zero as x goes to 1

Answer 18

oh but that is the whole point, you have to see if they match up!

Answer 19

limit only exists if its the same number from the right as it is from the left ....

Answer 20

what amistre said.
\[[x]\] is tricky at integers. it has a jump there so your real job is to check limit from right and left and see if they agree.

Answer 21

now come to 2ns question f(x)= x-[x] for all x belongs to real no... and in that question we have to find points of discontinuity

Answer 22

they will match because if
\[0<x<1\] you have
\[[2x]=1\] so your function is
\[x-1\] which also has a limit of zero as x goes to 1

Answer 23

i sent you a picture of the second problem. you can see it is discontinuous at all integers

Answer 24

i need a solved problem.. i am unable to understand from pic

Answer 25

what does "discontinuous" mean? it means there is a jump in the graph, so picture gives it. but lets use some algebra if that would help

Answer 26

yes solve it by algebra

Answer 27

lets look at
\[x-[x]\] on the interval [0,2)
for
\[0\leq x <1\] you have
\[[x]=0\]

Answer 28

so on the interval
\[[0,1)\] you have
\[f(x)=x-[x]=x-0=x\]

Answer 29

on the interval
\[[1,2)\] you have
\[[x]=1\] so on that interval
\[f(x)=x-[x]=x-1\]

Answer 30

and so one. so on the interval
\[[2,3)\] you have
\[f(x)=x-2\] etc etc

Answer 31

in other words you have a piece wise function. on the interval
\[[n,n+1)\] you have
\[f(x)=x-n\]

Answer 32

this is clearly not continuous for any integer n because the limit from the left and from the right do not agree

Answer 33

ok.thanks

Answer 34

:)

Answer 35

if x < n you have
\[\lim_{x\rightarrow n^-}f(x)=\lim_{x\rightarrow n-} x-(n-1)=1\] whereas if

Answer 36

now what will be the limit of x->0 x[1/x]

Answer 37

and if x > n you have
\[\lim_{x\rightarrow n^+}f(x)=\lim_{x\rightarrow n+} x-n=0\]

Answer 38

there all gory details worked out

Answer 39

you want a guess? i would say
\[\lim_{x\rightarrow 0}x[\frac{1}{x}]=1\]

Answer 40

yes...but i think it should be zero as we have to put 1 in [1/x] and the it will be x and the put 0?

Answer 41

but guess what you have to do? take the limit from the left and from the right. that is the whole job here i am sure. see that you get one in any case

Answer 42

you lost me. you are taking the limit as x goes to zero, not as x goes to one

Answer 43

so limit goes to zero ... as x=0 xpression will be 0[1/0]..

Answer 44

this is what you have to check, that
\[\lim_{x\rightarrow 0^+}x[\frac{1}{x}]=\lim_{x\rightarrow 0^-}x[\frac{1}{x}]\] and my guess is that they do, and that you get 1 in both cases

Answer 45

by putting 0 both goes to infinity

Answer 46

another good point to bring out is that the value of the limit does not care about what the actual value of the function is at the point.

Answer 47

you cannot compute this limit by replacing x by 0 because you will get 0/0 which is meaningless. you have to do something else. also you have to get rid of the greatest integer symbol and write the piecewise version of this thing

Answer 48

right, what amistre said. this thing has no value if you replace x by zero. you have to find the limit.

Answer 49

ok ..thank u all..take care ...:)

Answer 50

good luck. btw i am sure the answer is 1