lim x-->0 {cot x - 1/x}^n?
409386
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
807227
Take the derivative of top & bottom of the fractions, & put them in place of the original values, then set x value to the limit. \n\nlim x/sin(x) = lim 1/cos(x) = 1  (all limits as x → 0) \n\nlim (1-cos(x))/x = lim sin(x)/1 = 0 (all limits as x → 0)\n\nlim (4 sin2x)/(5 sin3x) = lim [(4×2cos2x+0)/(5×3cos3x)] = 8/15 (all limits as x → 0)\n\nIf the rule gives a result that is still 0/0 or ∞/∞ in the limit, apply it again…\n\nlim [(4x²−3x+2)/(7x²+2x−1)] = lim [(8x−3)/(14x+2)] = lim (8/14) = 8/14 (all limits as x → ∞ )
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
914435
x > 5\n\n- - - - - -s-
1023164
Limit ( n to infinity) (n + i/n)^ n
1065775
If a(N)=Sum(r^n, 0=>N) and -1<r<1...\n\n Is something that looks like 1/(1-r^N)
879179
There is no limit.\n\nYou can see this by approaching from two different directions, along the x-axis and along the y-axis.If x is zero and y->0, then the formula becomes:\n\n   (siny)^2/y^2\n\nSince sin y/y -> 1 as y->0, this also approaches 1 as y approaches zero.\n\nOn the other hand, if y=0, and x->0, then the formula becomes:\n\n    x^2/(2x^2)\n\nwhich has a limit of 1/2 as x->0.
1048409
It can be anything.  If you follow the sequence to (x-x) you will get an answer of 0 for the entire equation.
538748
Yes, but you need to account for two cases. The full solution:\n\n(x-1) / (x-10) < 0\n\nx-1 < 0 [if x > 10] OR\nx-1 > 0 [if x < 10] (by "flip the sign" rule; x cannot be 10)\n\nFirst inequality can never be satisfied.\nSecond inequality means 10 > x > 1.\n\nYou do not use foil for that.\n\nClassical (and Kemmy) probably interpreted it as a number/phase line, which can be faster in a simple case like this. For more difficult inequalities it's best to write out steps.
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
780025
x ≤ 5\n\n-- - - - - - -s-
1081409
I assume you mean:\nlim n->infinity of (1+n)^(1/n)\nIn this case, the limit is one.\n\nHowever, if you do:\nlim n->infinity of (1+(1/n))^n\nit is equal to e, or 2.718.....\n\n:)
308163
x can be any value.\n\ntan x = sin x / cos x\n\ncot x = cos x / sin x\n\nso tan x * cot x = sin x/cos x * cos x/sin x\n\nsin x's cross out each other\n\ncos x's cross out each other\n\nso you're left with 1.
712410
limit theta tending to 0 sintheta/theta =1
910686
lim x →∞ [x ^ ln x / ln x ^ x]\n= lim x →∞ [x^ ln x / xln x]\n= lim x →∞ [x^ (ln x - 1) / ln x]\n= lim x →∞ [e^{ln x * (ln x - 1)} / ln x]\n= lim x →∞ [{1/x(ln x - 1) + ln x*1/x}*e^{ln x * (ln x - 1)}/1/x] by l'Hopital's Rule\n= lim x →∞ [(2ln x - 1) * e^{ln x * (ln x - 1)}\n= lim x →∞ [(2ln x - 1) * x^ln (x/e)]\n= ∞\n\nPlease ignore as according to Excel this is clearly incorrect!!\nThe limit = 0\n\nMy error arose from interpreting lnx ^ x as ln(x^x) rather than (ln x) ^ x\n\nLet x = e^p. Then lnx = p\nlim x →∞ [x ^ ln x / ln x ^ x] Let x = e^p Then lnx = p\n= lim p →∞ [e^p²/p^(e^p)] and as x→∞, p→∞\n= lim p →∞ [e^p²/e^(lnp*e^p)]\n= lim p →∞ [e^(p² - lnp.e^p)]\nExamining the index p² - lnp.e^p\nLim p →∞ (p² - lnp.e^p)\n= Lim p →∞ (2p - 1/p.e^p - lnp.e^p)\n= -∞\nSo lim p →∞ [e^(p² - lnp.e^p)] = 0
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