limit(((n!)^(1/n))/n),n=infinity?
613021
http://www.ask.com/web?q=limit%28%28%28n%21%29%5E%281%2Fn%29%29%2Fn%29%2Cn%3Dinfini...+&qsrc=0&o=0
891612
You can convert f(x) as follows:\n\nf(x) = 1 / SQRT(4*(1+x/4)) = 1/[SQRT(4) * SQRT(1+x/4)]\nf(x) = 0.5 * 1/(1+x/4)^(1/2) = 0.5 * (1+x/4)^(-1/2)\n\nUsing the approximation:\nf(x) ~ 0.5 * (1 - x/8) = 0.5 (when x--> 0)\n\nThe only thing I don't understand is why you need the approximation.  The limit can be calculated straightforward without it.
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
104840
Actually, as what I have learned, this equation is limitless because the numbers itself are limitless. The incredible fact about this is that in every number you substitute in "x", the answer is always "0".
1165584
Thats a common misconception. \n limit x tends to 0,    5/x is infinity\nbut tends to 0 is not that x is equal to 0.It means x is almost 0(you are moving towards 0) but not 0.\nU cannot divide bu 0,its a undefined proposition.
72221
This is just a sneaky way of getting back to your question\non "0!".\n    This is the real answer.\n The definition of n! is a recursive one. It has a recursive\npart and a base. The recursive part is:\n                n!=n*(n-1)!\nThe base is:    1!= 1\nSubstituting 1 for n in the recursive part we get\n                1!=1*(1-1)!   The left side is defined as 1 and simplifying the right side we get\n                1=1*0!  and of course 1 times anything is\nitself so we get  1=0!  \n     Just thought you should know.
932401
the answer is 0\n\nu know, lim 1/(n+1) = 0\n              n -> ∞ \n\nand, lim (n/n+1)^n = 1\n        n -> ∞ \n\nso: lim (1/n+1)(n/n+1)^n = lim (1/n+1) * lim(n/n+1)^n = 0 * 1 = 0\n      n -> ∞                           n -> ∞          n -> ∞
98782
Just substituting x for 0 yields -9/-3 = 3.  This is not the indeterminate form 0/0 hence the limit is obtained by substitution, ie the answer is 3.
951210
ok...well 1/x becomes basically infinity as x gets really close to zero...so (1/x)/x woud be infinity/something really close to 0 which would make the value infinity\n\nso we have\n\ne-(1-x)^infinity\nwell...thats ok\n\ne-(1 - (basically zero))^infinity is\ne-1^infinity\nwhich is\ne-1\n\nso the limit as x approachs zero from the right is e-1 which is approximately 1.71828\n\nhope this helps\nmatttlocke
358921
(9^9)!\n\ni.e. factorial of 9 to the power 9\n\n\n\n\n\n\n--- ► λ ◄
812449
1/n! - 1/(n+1)! =  (n+1)!/[(n!)(n+1)!] - n!/[(n!)(n+1)!]\n                        = 1/[(n!)(n+1)!] X {(n!)(n+1-1)}\n                        = n*n!/[(n!)(n+1)!]                                 (divide by n)\n                        = n!/[(n-1)!(n+1)!]                  (divide by n!)\n                        =  1/[(n-1)!(n+1)]
682438
Infinity.Any number divided by 0 is infinity.
4300
zero divided by zero is not defined\none divided by zero is infinity\n\nIf you have a series like a_n/b_n whre both and b converge to zero as n goes to infinity, it might still have a limit, determined by what goes to zero faster a and b
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
208997
The only way to get the expression infinity/-infinity is through a limiting process in the real numbers, so I'll assume you're talking about limits. In that case the form above is indeterminate, meaning that it depends on the functional forms in the limit. \nSuppose you have f(x) and g(x) such that lim [f(x)/g(x)] = infty/-infty. Consider the equivalent expression lim [f(x)/g(x)] = lim [(1/g(x))/(1/f(x))] which approaches the form 0/0. This allows the use of L'Hospital's rule to evaluate the limit, which we know from experience can be anything.
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
377994
Is this (x + 3)/[x*(x - 1)^2] ? In that case, the limit is as you stated, positive infinity. \nA limit is never undefined; it either exists, or it does not exist. In the special case that it does not exist and it increases without bound or decreases without bound from all approaches, we say the limit is positive or negative infinity respectively.\nAs always, when a limit diverges, look at the behavior of the expression in an arbitrarily small region around the limit point (in this case, x=1). Since the denominator is a square multiplied by x, it is never negative in a sufficiently small neighborhood of 1 (specifically, any neighborhood contained in the interval (0,infinity) ). Similarly, the numerator is never negative in a sufficiently small neighborhood of 1, so the behavior of the whole expression is to diverge to positive infinity, no matter how it is approached.