What is lim(e^(4x)-1 -4x)/x^2 when x approaches infinity?
916864
Ord(e^(4x)-1 -4x) ~ Ord(e^(4x) > Ord(x^2)\nThen:\n\nlim(e^(4x)-1 -4x)/x^2 {x--> oo} = oo\n\nRemember\n\nln x < x^a < e^x < x^x
108152
(1/3)*e^(3x)-4sinx + c, where c is any arbitrary constant.
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
92912
I'm on a roll working with the definition, so I\nthought I'd try this one too. \n\nf(x)=4x/(5x+2)\n\nSubstitute it into the derivative definition:\n\nf'(x)=lim h->0 ((4(x+h)/(5(x+h)+2)) - 4x/(5x+2))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((4x+4h)/(5x+5h+2) - (4x/(5x+2)))/h\n\nCombine the fractions:\n\nf'(x)=lim h->0 (((4x+4h)(5x+2)-(4x)(5x+5h+2))/((5x+5h+2)(5x+2)))/h\n\nMultiply the adjacent terms:\n\nf'(x)=lim h->0 ((20x^2+20xh+8x+8h-20x^2-20xh-8x)/(25x^2+10x+25xh+10h+10x+4))/h\n\nSimplify:\n\nf'(x)=lim h->0 (8h/25x^2+20x+25xh+10h+4)/h\n\nCancel out the "h" from the top and bottom:\n\nf'(x)=lim h->0 8/25x^2+20x+25xh+10h+4\n\nSince h being zero won't cause an infinity, take the limit:\n\nf'(x) = 8/25x^2+20x+4\n\nBut notice that the denominator can be factored:\n\nf'(x) = 8/(5x+2)^2\n\nYou might notice that this equal to answer #1,\nexcept that it is reduced differently.
573378
as writen: \n8+1/3, 8+2/3, 8+1/4, 8+2/4, 8+1/5, 8+2/5, 8+1/6, 8+2/6\n\nlooks like it will converge to 8\nif you continue adding 1/x and 2/x where x can be any real number. eventually x will become significantly large compared to 1 and 2 that it will approach 0.\n\nin other words:\nthe limit of 1/x as x -> infinity = 0\n\n\n\nthe second looks like it converges to a limit of 6
1023164
Limit ( n to infinity) (n + i/n)^ n
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.
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.
1048224
This question is really only interesting if what you MEANT to type was:\n(4 - x^2) / (x + 2)\n\nSo you want to solve:\n\n. lim (4 - x^2) / (x + 2)\nx→ -2\n\nOf course the original expression is not defined at x = -2, because that would involve dividing by zero. However, for any real numbers x ≠ -2:\n(4 - x^2) / (x + 2) = (2 - x)(2 + x) / (x + 2)\n\nSince x ≠ -2 it is valid to divide top and bottom of the fraction by the common term (x + 2) which leaves:\n\n(4 - x^2) / (x + 2) = (2 - x), for all x ≠ -2.\n\nThus,\n. lim (4 - x^2)/(x + 2)\nx→ -2\n= lim (2 - x) = 2 - (-2) = 4 <<<<<<<<<<<\n. x→ -2
744013
This is not common.\n\nIn general, you're looking for x^y = y^x.\n\nThis is equivalent to x^(y/x) = y, which in turn is equivalent to x^(1/x)=y^(1/y).\n\nx^(1/x) has a maximum at e^(1/e) = 1.44466....  The function is increasing before that point and decreasing after.  As a result, for each number there is another on the opposite side with the same value.  2^(1/2) = 4^(1/4) <--> 2^4 = 4^2 is the only example of this in the integers.
12883
how do you count to infinity?
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.
614800
It converges to one.\nNote that |x| = x if x > 0 and |x| = -x if x < 0.\nBreak the integral into two pieces.\nint from -infinity to 0 of -xe^(-x^2) dx \n+ int from 0 to infinity of xe^(-x^2) dx\nUse substitution letting u = -x^2\nint from -infinity to 0 of (1/2)e^u du\n+ int from 0 to -infinity of (1/2)e^u du\n= (1/2)e^u from -infinity to 0 -(1/2) e^u from -infinity to 0\n= (1/2)[e^0 - 0] - (1/2)[0 - e^0]\n= (1/2)(1) - (1/2)(-1)\n= 1\nI don't have this typed quite correctly.  You have to replace -infinity with b and then take the limit as b goes to -infinity.
797160
Your function y=4*x^2*e^(5x) is defined for any x. Meanwhile x^2>=0 & e^(5x)>0, thus y>=0 for any x. Now y’=4*(2x+5x^2)*e^(5x), local extrema being at y’=0. Or x1=0, x2=-0.4\nHence and “above” x1=0 means local min & y(0)=0, so x2=-0.4 means local max, \nLim[for x->-infinity]=0 & Lim[for x->+infinity]=+infinity. Does it help you Xu?
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
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0