what is the limit of  : [ x [1/x] ]   when x goes to zero?( if it does not have limit , tell me why ? )?
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
1148883
I assume what looks 'funny' to you is the wild oscillations near x=0.  The reason is that for x near zero, the quantity 1/x gets really large, and as x gets closer to zero the value of 1/x changes a LOT for a small change in x.  What this means for sin(1/x) is that the sine function goes through many oscillations over just a small range of x (near zero) because its argument changes so much.  Does that make sense?\n\nSee if you can make an argument for what behavior to expect for sin (1/x) when x is far from zero instead (positive or negative).
270868
well, you just multiply top and bottom by x, like how you simply fractions.  (clearly x isn't 0 so this is legal)\n\n= lim (x->1) x(1-1/x)/(x(x-1))\n= lim (x->1) (x-1)/(x(x-1))\n= lim (x->1) 1/x \n= 1
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.
252229
Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
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".
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
981709
To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
168972
Your question does not make sense. x cannot be less than or equal to 0 and less than or equal to 7 unless it is less than or equal to 0, which I do not think you mean. \n\nI think you mean 0<=x<=7\n\nThe rest of the question makes no sense. \n\nAnd it really smells like a homework problem. Try thinking for a change.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
1055950
Yes, linear functions have limits.\n\nFor example, the limit of 3x+2 as x approaches 4 is 14. Here's how to prove it:\n\nFor every e>0, choose d=e/3. Then, if 0<|x-4|<d, we have\n|(3x+2)-14|=|3x-12|=3|x-4|\n<3d=e.\n\nFor this situation, we say that the limit of a function f(x) as x p&#92;approaches a is L is for every tolerance e>0, there is a tolerance d>0 so that whenever 0<|x-a|<d, we have\n|f(x)-L|<e.
936881
There is a local maximum at x = c on f(x) if\nf'(c) = 0 AND f"(c) < 0\n\nOccasionally also when f"(c) = zero .. this occurs when there is a multiple root at that value of x.\nThen further testing is needed. In this case:\n1. if If f(c-) < f(c) AND f(c+)< f(c) then it is a local maximum;OR\n2. if f'(c-) >0 and f'(c+) <0 then it is a local maximum; OR\n3. if f"(c-) <0 and f"(c+) <0 then it is a local maximum
289224
(x+ [1+i*sqrt(3)]/2) * (x+[1-i*sqrt(3)]/2) = 0
793364
To show that it doesn't exist, look at the path where x=0 and y moves to zero.  The problem reduces to:\n\nsin(y)/y^2\n\nYou can use L'Hospital's rule to show that this limit does not exist.
727467
It can be a constant or a variable.  It asks what is the behavior of the function as x approaches the "a" variable or constant.  Thus,\nlim x->a [(x - a)²]/[1 + cos (π/a x)]\n\nLet me show you.  We first get the limit:\n_______________________________________\n| Since it is the indeterminate form 0/0, we can use the \n| L'Hopital's rule.\n| lim x->a [2x - 2a]/[- π/a sin (π/a x)]\n|\n| Again, it is 0/0, so use L'Hopital's Rule,\n| lim x->a 2/(-π²/a² cos (π/a x)]\n|\n| We can now substitute x\n| = 2/(-π²/a² cos π)\n| = 2/[-π²/a² (-1)]\n| = 2/(π²/a²)\n| = 2a²/π²\n|__________________________________\n\nWe now have the limit\n= 2a²/π²\n\nNow, you can interpret this limit in 2 ways.\n\nIf "a" is a constant, then the limit of that function (of only 1 variable, x) when the variable x approaches an unstated constant "a" is the limit that we have acquired (which, in this case, is a constant).\n\nNow, if "a" is a variable, then the limit of that function (of 2 variables, a and x) when the variable x approaches the value of the variable a, is a function of a single variable, which is the limit we have acquired (which, in this case, is a function).\n\n^_^