What is the limit (calculus)?
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If you are permitted to use L'hopital's rule, these are simple. Substituting, we find that both of these limits yield the indeterminate form 0/0, so we have:\n\n[x→0]lim sin (-2x)/x = [x→0]lim -2 cos (-2x)/1 = -2\n\n[x→0]lim sin (x+π)/x = [x→0] cos (x+π)/1 = -1\n\nIf you have to evaluate these without L'hopital (which I suspect, if you can't figure them out), then this is slightly harder, but still possible. Taking the first one:\n\n[x→0]lim sin (-2x)/x\nUsing sin (-x)=-sin x:\n[x→0]lim -sin (2x)/x\nApplying the double angle formula:\n[x→0]lim -2sin x cos x/x\nRearranging:\n[x→0]lim -2 cos x sin x/x\nWe know that [x→0]lim sin x/x = 1 (this may be proven geometrically, without L'hopital, and in fact is an integral part of the proof that d(sin x)/dx = cos x). Thus, considering sin x/x as a single unit, this limit assumes the determinate form -2 * 1 * 1, so:\n[x→0]lim sin (-2x)/x = -2\n\nFor the second:\n[x→0]lim sin (x+π)/x\nUsing the angle addition formula:\n[x→0]lim (sin x cos π + cos x sin π)/x\nSimplifying:\n[x→0]lim -sin x/x\nAnd because [x→0]lim sin x/x = 1:\n[x→0]lim sin (x+π)/x = -1
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.
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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
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
1154919
<1>...u know lim xsin(1/x)=lim (sin(1/x))/(1/x),and when x goes to infinity ,(1/x) goes to <0>....and we know that (sina)/a when <a> goes to <0> is 1...and here <a> is <1/x>...so it's limit is <1>
819082
Let f(x) = x ³/ (x² + 2x - 15) \n\n= x - 2 + (19x -30)/(x² + 2x - 15) (by division)\n\n= x - 2 + (19x -30)/[(x - 3)(x + 5)]\n\nSo f(x) is undefined when x = -5, 3\n\nAlso lim x→±∞ f(x) = x - 2\n\nThus f(x) is asymptotic to g(x) = x - 2 as x →±∞ and asymptotic to x = -5 and x = 3
932401
the answer is 0\n\nu know, lim 1/(n+1) = 0\n              n -> ∞ \n\nand, lim (n/n+1)^n = 1\n        n -> ∞ \n\nso: lim (1/n+1)(n/n+1)^n = lim (1/n+1) * lim(n/n+1)^n = 0 * 1 = 0\n      n -> ∞                           n -> ∞          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.
804449
There is no limit. The problem is that sin(x) goes between postive one and negative one infinitely many times as x gets small (i.e. goes to negative infinity). The hard part, though, is to figure out how to show this.\n\nI will look at two different sequences of numbers that go to negative infinity. When I plug the numbers in the first sequence into x/(2sin(x)) they will go to positive infinity. When I plug the numbers in the second sequence into x/(ssin(x)) they will go to negative infinity. If I can get two different answers then there is no limit.\n\nThe first sequence is x_n = -pi/2, -pi/2-2pi, -pi/2-4pi, -pi/2-6pi, ....,-pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-pi/2-2n pi)/(2sin(-pi/2-2n pi))=(-pi/2-2n pi)/(-2)=pi/2+2n pi\nAs n gets big then this number goes to positive infinity.\n\nThe second sequence is x_n = -3pi/2, -3pi/2-2pi, -3pi/2-4pi, -3pi/2-6pi, ....,-3pi/2-2n pi\n\nWhen I plug this into x/(2sin(x)) I get\n(-3pi/2-2n pi)/(2*sin(-3pi/2-2n pi))=(-3pi/2-2n pi)/(2)=-3pi/2-2n pi\nAs n gets big then this number goes to negative infinity.\n\nWe now have two sequences of numbers that give different answers. The function cannot converge to a single number.
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.
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You have two cases.\nCase 1:\nIf c is negative, then e^(cx) goes to 0 as x goes to infinity.  The limit would then be zero.\nIf c is positive, then you can apply L'Hopital's rule.\nIt becomes\n[bce^(cx)/(a+be^(cx))]/d = bce^(cx)/[d(a+be^(cx))]\nThis also goes to infinity over infinity.  Use the rule again.\n[b(c^2)e^(cx)]/[dbce^(cx)]\nYou can cancel b, c and e^(cx) to get\nc/d\nAnswer:\nIf c < 0 , then the limit is 0.\nIf c > 0, then the limit is c/d.
711149
Is it lim (x approaches 0)?
1000517
You probably meant to ask:  Can pi be expressed as a limit?\n\nYes, it can.  And in many different ways, via summations, sequences, and integrations.  Here are some of the most popular ones:\n\nsumm(n=0,inf) (-1)^n/(2n+1) = 1/1 - 1/3 + 1/5 - 1/7 + 1/9 - ... = pi/4\n\nprod(n=1,inf) [(n+1)/n]^[(-1)^(n-1)] = 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 * ... = pi/2\n\nsqrt(2)/2 * (sqrt(2+sqrt(2))/2 * (sqrt(2+sqrt(2+sqrt(2)))/2 * ... = 2/pi\n\ninteg(-1,1) sqrt(1-x^2)dx = pi/2\n\nMany more exist using a wide variety of methods and with varying levels of complexity.  Some of them are the basis for the algorithms which let computers calculate pi to a large number of digits.
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^_^
1079050
Calculus is a central branch of mathematics, developed from algebra and geometry. It is built on two major complementary ideas, both of which rely critically on the concept of limits. The first is differential calculus, which is concerned with the instantaneous rate of change of quantities with respect to other quantities, or more precisely, the local behavior of functions. This can be illustrated by the slope of a function's graph. The second is integral calculus, which studies the accumulation of quantities, such as areas under a curve, linear distance traveled, or volume displaced. These two processes act inversely to each other, a fact delivered conclusively by the fundamental theorem of calculus.\n\nExamples of typical differential calculus problems include:\n\nfinding the acceleration and velocity of a free-falling body at a particular moment. \nfinding the optimal number of units a company should produce to maximize their profit. \nExamples of integral calculus problems include:\n\nfinding areas and volumes \nfinding the amount of water pumped by a pump with a set power input but varying conditions of pumping losses and pressure \nfinding the amount of parking lot plowed by a snowplow of given power with varying rates of snowfall. \nToday, calculus is used in every branch of the physical sciences, in computer science, in statistics, and in engineering; in economics, business, and medicine; and as a general method whenever the goal is an optimal solution to a problem that can be given in mathematical form.\n\nFundamental theorem of calculus\nMain article: Fundamental theorem of calculus \nThe fundamental theorem of calculus states that differentiation and integration are, in a certain sense, inverse operations. More precisely, if one defines one function as the integral of another, continuous, function, then differentiating the newly defined function returns the function you started with. Furthermore, if you want to find the value of a definite integral, you usually do so by evaluating an antiderivative.\n\nHere is the mathematical formulation of the Fundamental Theorem of Calculus: If a function f is continuous on the interval [a, b] and if F is a function whose derivative is f on the interval [a, b], then\n\n \nAlso, for every x in the interval [a, b], \n \nThis realization, made by both Newton and Leibniz, was key to the massive proliferation of analytic results after their work became known. The fundamental theorem provides an algebraic method of computing many definite integrals—without performing limit processes—by finding formulas for antiderivatives. It is also a prototype solution of a differential equation. Differential equations relate an unknown function to its derivatives, and are ubiquitous in the sciences.
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L'Hospita;'s rule shall not apply here as it is non zero/zero form\n\nit applies for form 0/0 or infinite.infinite\n\nlimit f(x)/g(x) when it is of above form is lim f'(x)/g'(x) \n\nfor example as x->o lim sin x/ x is of the form 0/0\n= lim x->0 d/dx(sin x)/d/dx(x)\n= cos x/ 1= cos x = 1 when x->0\n\nfor your case it is 1/0 or infinite (no l'Hospital's rule)
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How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.