Problem 5RE

Limits of sequences Evaluate the limit of the sequence or state that it does not exist.

STEP_BY_STEP SOLUTION Step-1L’Hopital’s Rule for zero over zero ; Suppose that lim f(x) = 0 ,limg (x) = 0 , and that functions ‘f’ and ‘g’ are differentiable on an xa xa 1open interval ‘I Containing a. Assume also that, g (x) = / 0, in I , if x / a then ; 1 lim f(x) = lim f (x). xa g(x) xa g (x) so long as the limit is finite, +, or -. Similar results hold for x and x.L’Hopital’s Rule for infinity over infinity;suppose that functions ‘f’ and ‘g’ are differentiable for all x larger than some fixed number. If lim f(x) = ,limg (x) = then ; xa xa 1 lim f(x) = lim f (x). xa g(x) xa g (x) so long as the limit is finite, +, or -. Similar results hold for x and x. Step-2 n 1/n Given sequence is ; a n n = n It is an exponential form , here the base and power contains ‘n ‘ . So , in this casestake ‘ln’ on both sides .Then the given sequence becomes; Let , f(x) = n 1/n ln( f(x) ) = ln( n 1/n )\n 1 m = n ln(n) , since ln(a ) = m ln(a). lim 1 ln(n) = lim ln(n. n n n n ln(n)is a rational function , let P(n) = ln(n) , Q(n) = n n As , n tends to infinity , P(n) also tending to infinity , and as , n tends to infinity , Q(n) alsotends to infinity. So , in this cases we can use L’Hopital’s Rule 1 P(x) P (x) That is , xa Q(x) = xa Q (x) . ln(n) d(ln(n)) Therefore , lim = lim d(n) n n n dn 1 = lim n , since d ln(x) = 1/x. n 1 dx = lim n. n = 0 , since as n , then 1 . n 1/n Therefore , lim ln(n ) = 0. n lim n1/n = e 0 = 1, since ln(a ) = b ,that implies a = e b n Therefore, lim a = lim n1/n = 1 n n n