23my wrong , i thought u were telling something else
especially to ppl who think they r genius ...
like
1) shubomoy
2)sanchit
3)जय
4)qwerty
prove :
\boxed{1} \\ \texttt{{if |x| < 1 , then prove that }} \\ \\ \lim_{n\rightarrow \infty}x^n=0 \\ \\ \boxed{2}\lim_{n\rightarrow \infty}\frac{x^n}{n!}=0
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17 Answers
62yaar. sorry to not agree to you on this one..
not that i am taking a soft stand for him... i dont want to say things about individual students and i do not think it is just subhomoy...but even the other guys..
I think you should realize that this is a public forum and that you will meet all kinds of people.
If there are not so good people whom you you dont like you can simply ignore them and not be one of them by talking low about them...
1dear nishanth sah sir ,
i know that Mr. shubomoy is ur student u have reservations for him and he succesfully extraxted that sympathy
in my dictionary
i define
GOD --- kaymant sir , prophet sir , nishanth sir
but this guy(shubomoy) has been consistently showing his attituide
let me remind even people like celestine preetham , rohan , AP , etc were humble
i am sure u will defy my statements
but these are personal opinions !!!
62no doubt this one was to offend.. but dont mind subhomoy.. atleast he has not put prophet sir's or kaymant sir's name here [1][3][4]
1i posted one question and u fell in love ...
better will be to do open a das gupta and solve sums
and come here less often !!!
now u have got addicted to the website ...so bad nothing can be done now
1u can proceed the first one taking
|x|=\frac{1}{1+b} b\in (0,\infty)
now apply binomial expansion and squueze theorem
1johny is proof is mathematically rigorous and correct
other solution are ' intuition '
39utkarsh
good work for seeing in the case of integers
but u see, here we just not only consider integers but all the real numbers that satisfy the givven condistion
like in the first question, its given |x|<1
we have infinitely mane numbers less than one which are not integers u see like 0.123555555555555 , 0.345666666666666 and so on
so a proof which takes into consideration all these numbers is required in this case
nevertheless what u have said is correct in the case of integers
1hey i am not a great mathematician or a genius. i study in class 9. i want to know whether i am correct or not. just guide me. if |x| < 1 then it must be 0 if x belongs to integers. then it is obvious that 0n is always 0. and as in the second case, 0 / anything (in this case n!) is always 0. am i correct?
392)
let kn denote the n th term of the series given by x^n/n!
now we have
kn+!/kn=x/(n+1)
whenever we have n > x - 1,
kn+1/kn is less then 1, or kn+1=lnkn where ln satisfies the condition ln<1
so we find that value of kn decreases as value of `n` increases
hence for n tending to infinity, the limit is 0
23\frac{x^{n}}{n!}= (\frac{x}{1})(\frac{x}{2})(\frac{x}{3}).....(\frac{x}{n})
each term in the bracket is a fraction
as the denominator increases the fraction decreases and
\frac{x}{n}\rightarrow 0 ,as as n →∞
fraction x fraction = gives a smaller number. So on multiplying these decreasing infinite fractions , the limit tends to infintesimally small number , wich →0 as n→∞
donno why on earth i wrote \frac{x^{n}}{n!}=(\frac{x}{n})(\frac{x}{n})(\frac{x}{n})... earlier...