lim x -> 0\n\nof\n\n(e^x -1)^3 divided by (x-2)e^x + x + 2
129690
I had to use L'Hopital's rule three times.\nAfter the second time, make sure you simplify the denominator before you take the derivative a third time.\nEdit: You are right.  It is 6.
57818
D.    X (greater or equal to) 4\n      2(-2)+4 = 0\nso (-infin, -2) U [4, infin)
367961
x+78 [greater than or equal to] 120\nx [greater than or equal to] 120-78\nx [greater than or equal to] 42
674625
Lim { [ sin(x-1)  ] / [ x^2 + x - 2] }\nx-->1\n\nAdd the following to the denominator\n-2x + 2x +1 -1\n\nLim { [ sin(x-1)  ] / [ x^2 + x - 2 -2x +2x +1 -1] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ x^2 -2x + 1 +3x -3] }\nx-->1\n\nNote: (x^2-2x+1) =(x-1)^2    and  (3x-3) = 3(x-1)\n\nThen\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)[ (x-1)+3 ] }\nx-->1\n\nLim { [ sin(x-1) / (x-1) ] * [ 1/ (x-1+3 )] }\nx-->1\n\nLet u= x-1 then as x-->1, u--> 0\nand\nLim { [ sin u / u ] * [ 1/ (u+3 )] }\nu-->0\n\nSince\n\nLim  [ sin(u) / (u) ]=1\nu-->0\n\nThen \nLim { [ sin u / u ] * [ 1/ (u+3 )] } = 1 * [1/(0+3)] = 1*1/3 = 1/3\nu-->0
636222
err unless i'm misreading the question, then the answer is undefined.  the denominator is 0^3 which is 0.  you can't divide by 0.
477040
f(x)=x^3+5x^2+3x+1\nf'(x)=3x^2+10x+3\nsetting it to zero\n(3x+1)(x+3)=0\nx=-1/3 and x=-3\nf''(x)=6x+10\nat x=-1/3 f''(x)=+ve and so min\nat x=-3 f''(x)=-ve and so max
181499
lim x-->0 {cot x - 1/x}^n = (lim x-->0 {cot x - 1/x})^n\nBut lim x-->0 {cot x - 1/x}= lim x-->0 {cos x/sin x - 1/x}\n = lim x-->0 ({ x cos x - sin x )/(x sin x))\nDifferentiating numerator. & denominator \n =lim x-->0 ((-x sin x+ cos x - cos x )/( x cos x + sin x))\nDividding numerator. & denominator  by x\n =lim x-->0 ((- sin x )/( cos x + (sin x)/x)) \n = 0/(1+1) = 0\nSo lim x-->0 {cot x - 1/x}^n = 0^n = 0
529296
Let x be one number\nother number=100+x\nproduct\n=x[100+x]\n=100x+x^2\ndifferentiating\nf'x=100+2x\nthis is a minimum \nwe put this=0\n100+2x=0\nx=-50\nthe numbers are\n50,-50
174505
haha, but this does not work. you cannot mathematically divide by (a-b) because this quantity is equal to 0. (1-1)=0. Dividing by zero yields an indeterminate answer.
537675
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
88474
f(E) = 1/(exp{(E-Ef)/kT}+1)
484490
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.
483551
You divide the bottom into the top and then see what happens as the limit of x--> infinity.\n\nso x^3/(x^2-4) = x+4x/(x^2-4). Now notice that when x goes to infinity the second part of our answer is going to 0 because the x^2 part is going to dominate the 4x in the numerator. So when this function gets very large it is going to be closer and closer to y= x +0.\n\nThis is all that is meant by asymptotes. They are the functions that aproximate another function as x reacher infinity or a certain constant where the equation is undefined (0 in the denominator)\n\nhope this helps.
388563
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.
544894
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
103664
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.
409704
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless