[[x]]+[[-x]] what are the limits if they exist and where is it discontinuous?
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".
651114
it would be X<12, with extra line
665272
To build on what "Thin Kaboudit" was saying, you can take the limit of a function f(x) = 3/x as x goes to zero.\n\nLets say you decided to approach x from the positive side, slowly decreasing its value toward zero.\nThe closer and closer x gets to zero, the closer the function's value gets to [positive] infinity.\nNow lets say you decide to approach x from the negative direction, starting out with negative numbers and slow increasing them toward zero.  As you get closer and closet to zero from the negagtive side, the function goes toward negative infinity.\n\nSince the limit from the positive side does not equal the limit from the negative side (infinity does not equal negative infinity), the limit of the function does not exist since the function is not defined when x = 0.
211910
(-3)-[(-2)-(4)(-5)] \n(-3)-[(-2)-(-20)]\n(-3)-[-2+20]\n(-3)-18\n-21
867867
Ignore the relative maximum from -2<x<3. The horizontal asymtote is defined to be the limit as x approaches infinity of the function. Which as you have calculated is 3.
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
1146893
x^2+mx+1/x^2+x+1<3\n\ncase 1 : for x>0\n=> (x+1/x)^2 +(m+1)x <4\n=> (m+1)x < 4- (x+1/x)^2\n\nNow, we find the maximum value of the RHS:\nthe minimum value of (x+1/x)^2 is 2 when x=1 and for all other values of x it is >2 \nThis makes max(RHS)=0 or m+1 <0 \nor m<-1\n\ncase 2: x<0\nthe same argument as above(but x=-1)  leads us to m+1 >0 \nor m>-1\n\ncase 3: x=0\nThe given expression is undefined \n\nI guess its a home work problem, which indeed makes you lazy!
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
1047475
U r correct. the end points in a closed interval can be the value of c because c belongs to [-1,1].
400503
since each one gives you a *-----* i would say all of them are line segments.\n\nHowever since D. is the only one with an indefinite beginning and end and not a possible end or beginning, i would definately have to go with \n\nD.) -5 <= x <= 5\n\nThat is the only one that has a defined beginning and end and not a undefined end and/or beginning.
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.
857286
Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
1149923
(2,-4) if f(x) is defined for x=2
854009
lim as x approaches 0 from the positive of (ln x) is negative infinity.\nlim as x approaches 0 from the positive of (sqrt x) is 0.\n\nif the numerator goes to negative infinity and the denominator goes to 0, the limit is negative infinity. it might look like L'hospital's rule is required, but it is not.
518111
As x->0, -3x->0. \nNow let -3x = z.\nThus the limit becomes\nLim z->0 [-3*{ln(1+z)}/z]\n= (-3)*Lim z->0 [{ln(1+z)}/z]\n= (-3)*1\n= -3\n\nSo 'c' is correct.
1183491
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
56574
They are x=-3 x=3 x=1 \n\nlim as x->3 from the right is infinity\nlim as x->3 from the left is negative infinity\n\nlim as x->-3 from the left is negative infinity\nlim as x->-3 from the right is infinty\n\nlim as x->1 from the left is 1/infinity\nlimit as x->1 from the right is -1/infinity