lim {[f(x)+x²}^1/{f(x)} =

x->0

 

a) eⁿ   b)e³  c)e²  d)e

 

 

 

 

 

_{[n][infinity ]} ⁿc_r(m/n)^r(1-m/n)^n-r =

 

 

a]e^-mm^r    b]m^r/r!    c]m^re^-m/r!   d]e^-rr^m/m!

 

 

 

 

_[x][0]  {(1-x)^1/x-e+ex/2}/x²=

 

 

 

 

             

_{[n][infinity]} (n!/nⁿ)^1/n=

 

 

 

a]1/n  b]n   c]e     d]1/e

 denoting infinity by i
 
  ANSWER TO  lim nCr p^rq^(n-r) =c] [e^(-m)( m^r)]/ r!
             n->i

  lim nCr p^rq^(n-r)
  n->i

  lim  n!/((n-r)!*r!) p^rq^(n-r)
  n->i
 
 now using np=m, p+q=1 where m ,r are  constants

 lim  n!/((n-r)!*r!) (m/n)^r*{1-(m/n)}^(n-r)
 n->i

 lim  n!/((n-r)!*r!)* [(m/n)^r*{1-(m/n)}^n )/(1 - ( m/n))^r
 n->i
 
 

 lim {1-(m/n)}^n = e^(-m)
 n->i

 when  n->i  { [n!/(n-r)!]/n^r} becomes  1

 therefore we get {e^(-m)*m^r}/r!

 

 

 

 ANSWER TO  lim ( n!/(n^n))^(1/n) =d] 1/e
            n->i 
 
          let y= ( n!/(n^n))^(1/n)
          log y = [log n! -nlogn]/n
 
 n!=n*(n-1)*(n-2)......3*2*1  
 log n!=logn +log(n-1) +log(n-2)+.......        
 log n= logn +logn+......ntimes 
 
 NOW USING THE FORMULA OF SERIES REPRESENTED BY DEFINITE INTEGRALS WHEN n TENDS TO INFINITY

lim  log y   = SIGMA log(1- (r/n)) WITH r varying from 0 to n
n->i                 n->i  

 which when converted gives
            INTEGRATION OF log(1-x) with limits from 0 to 1
                            where x=(r/n)
              INTEGRATION gives -1
 ANTILOG OF -1 IS 1/e { we take antilog as we have taken log y)