9no false ofcourse
Which of these is greatest
.9 (0.9999999999999999999................)
1.01 (1.00000000000000.................0001)
1
??
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70 Answers
62no celestine
is this true?
\lim_{x\rightarrow a}\left\{f(x) \right\} = f \left\{\lim_{x\rightarrow a}x \right\}
9.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999.................... lies close but to left of 1 in numberline
so its zero only
consider [x] graph now wat is the value of position just left of 1 ?????
another argument
limit h→0- [1 + h] = 0 now wat is 1+h- here ???
its .999999999999999999999999999999.......... only na ?
62celestine.. i am on the verge of giving up hope of explainingthis one to you :D
well how do you know that
.99999999....... < 1
your premise of this assumption is wrong!
hence your assumption that
[.99999999999.......] = 0 is wrong!
9contradiction
we all know that greatest integer fn is [ ] non decreasing
now apply [] for 1 1.00000000000...1 and .9999999999999999999.......9
well get 1 ,1,0 as ans contradicting they r equal
now bhaiya i want u to give me any one contradiction to the fact they are unequal
1well i m trying some bad resoning
acc to me all r equal to 1
so in my result %error=+-(.00000000000........................1)%
too small to matter
1oh the bar does matter !!!!!!.......................
oh k !!... then they all are equal !
1dude sunil thts wat i did...........hehehe same pinch.......i bet tht hurt....
11if the difference is taken as h and since it is too small thats why it approaches zero.
→ .999999999999bar = lim(1-h) as h approaches zero = 1
and
1.00000000001 = lim (1+h) as h approaches 0 = 1
therefore all three are equal........
1sry feeling vry sleepy will join later
one last try
acc to sandwhich theorem other 2 result can tend to 1
only guessing not using pen
goodnight
1oh ok the that difference is equivalent to h. so lim(1-h)=1 and lim(1+h)=1.......
62vishal u have given a good reasoning...
Can you think in terms of limits?
11Bhaiya i know that 1.00000000001 is greater than 1 but since there is a bar on 0000000 so the difference is too small and it will keep on decreasing....
otherwise the answer should be 1.0000000001
62yeah it is bar .....
I will edit the question to make it mroe obvious!
62hmm... but vishal and sunil..
isnt it that 1.0000000000001 > 1 for any number of zeros?
And Akand.. and Unique... isnt it that the difference is very small and will keep decreasing to the minimum??
11.0000000000000001 think soooooo
but i dono truth is it bar abov zerorz??
1i think its 1.00000000001 but i dint understand vishal..........hw r they equal???
11Really a very good question.....
me too think all 3 are equal.....
62yup Philip you are right in a way..
This was more about the limit thing.. that there are increasing number of zeroes :)
Hopefully the message I was trying to convey went across :)
1Undoubtedly I write the limit = 0, but that does not mean that x2 is exactly equal to 0 when x tends to 0, right? Does it not mean that x2 tends to 0 when x tends to 0? Does not the entire soul of limits rest on the phrase 'tends to'?
62what do you write in exams?
\lim_{x\rightarrow 0} x^{2} = 0
or?
\lim_{x\rightarrow 0} x^{2} > 0
1Looks like this topic is closed. But anyway, I had a doubt. Consider this:
If lim f(x) = φ
x→a
Then it means that as x tends to a, f(x) tends to φ, it is never exactly equal to φ.
Similarly, if lim (1-h) = φ
h→0+
then it means that as h tends to 0+, 1-h tends to φ, it is never exactly equal to φ.
So, if φ = 0.9, it means that φ tends to 1-, it is never equal to 1.
Infact, isn't this the essence of limits: the concept of tending to a value, but never being exactly equal to it? Is this reasoning not correct? Please help. I'm new to this forum. Thank you.
1@ indraneel
once youve proved 0.9bar=1 then the best method to prove 1.0(bar)1=1 is
0.0(bar)1=1-0.9(bar)
1 + 0.0(bar)1=1+(1 - 0.9(bar))
1.0(bar)1=1+0=1 [6]