uses of mathematical limit in daily life?
472291
You run a bank and a competing bank has a branch across the street.\n\nYou compound your interest annually. So you use the formula: A = P * (1+i)^n.\n\nThe bank across the street says, "We'll compound your interest every month. Their formulate is: A = P * (1+i/12)^(n*12), which gives the customer a little better deal.\n\nOkay, this means war. You start advertising weekly compound interest, or: A = P * (1+i/52)^(n*52), which is a slightly even better deal.\n\nThe other bank compounds daily.\n\nSo you decide to take it as far as you can go. You'll compound continuously. You're formula is:\n\nA = P * (1+i/x)^(n*x) as x approaches infinity. \n\nThe bank across the street can't beat that. Oh, there's no infinity button on your calculator, so how do you calculate savings deposits?\n\nBy running this formula through limit theory, you use the fact that (1+1/x)^x as x approaches infinity is e, and you get the following formula:\n\nA = P*e^(i*n) for continuous compounding.\n\nI don't know of any bank that calculates interest continuously, but there are at least two natural phenomenons that change at a continuous rate.\n\nThe first is population growth, such as the number of humans on the plant, or the number of bacteria in a jar.\n\nThe other is radioactive decay, which is often described as "half-life."\n\nBoth of these natural behaviors are defined by the continuous growth formula, and it cannot be derived without limit theory.\n\nBy the away, we pretty much define calculus using limit theory, but one could argue that calculus isn't part of daily life.
1016907
As you specified that limit as 'x' tends to (pi/4), I have changed the solution accordingly.\n\nLim  x--> (pi/4) [{sin(x) - cos(x)}/cos(2x)]\n= Lim  x--> (pi/4) [{sin(x) - cos(x)}/{cos^2(x) - sin^2(x)}]\n= Lim  x--> (pi/4) [{sin(x) - cos(x)}/{(cos(x) + sin(x))*(cos(x) - sin(x))}]\n= Lim  x--> (pi/4) [-1/{cos(x) + sin(x)}]\n=  [-1/{cos(pi/4) + sin(pi/4)}]\n=  -1/[{1/sqrt(2)} + {1/sqrt(2)}]\n= -1/[2/sqrt(2)]\n= -{sqrt(2)}/2 \n\nYou don't need to use L.Hospital's rule.
328117
Yes. This is a Frequently Asked Question. Following is my previous answer:\n\nThe real numbers can be defined using Cauchy sequences. Basically, the real numbers make a number line continuous by filling in all possible "holes". If you have a set of real numbers A, and that set is bounded above by some real number (there is a real number that is greater than every real number in A), then we define the set to be a real number that is the smallest real number greater than or equal to every real number in A. This is how one constructs the real number system, by defining every point on a line segment to be a real distance from the start of the line segment (thus all square roots, corresponding to hypotenuses of right triangles, have been incorporated into our number system). \nIn other words, every bounded (from above) subset of rational numbers corresponds to a real number. Now, rational numbers are defined as the set of all ratios of integers. Just like two ratios of integers can correspond to the same rational number (for example, 2/4 and 500/1000 are the same rational number), so can two bounded subsets of rational numbers correspond to the same real number. \nA common way of representing these sets is to use decimal notation. Decimal notation is an infinite series in powers of ten. Each partial sum of the series is taken to be an element of the set in study. A series is said to converge if the set of partial sums is bounded as described above.\nFor example, the number 0.333... is the decimal way of writing the series 0 + 3/10 + 3/(10^2) + 3/(10^3) + ... which corresponds to the set of partial sums {0, 3/10, 33/100, 333/1000, ...}. It is trivial to show that this set is bounded by 1, so this set corresponds to a real number. It can be shown by algebraic manipulation that this set actually represents the rational number 1/3.\nLet us look at the set corresponding to 0.999..., or {0, 9/10, 99/100, 999/1000, ...}. This set corresponds to a real number because each element is of the type (1 - (9/10)^n)/(1/10) = 10(1 - (9/10)^n) (it is a geometric sum; the expression can be derived after a moment's thought). Obviously, no power of any fraction less than 1 is greater than 1 since all such powers are less than the original number, so this is bounded by 1.\nWe hypothesize that 1 is the smallest number that bounds this sequence. Consider the difference 1 - L where L is the smallest number that bounds this sequence. By the field axioms (common algebra), if 1 - L = 0, then 1 = L. Now L is greater than every element of the set in question by definition. Thus, consider 1 - 10(1 - (9/10)^n) = 9^n/10^(n - 1) - 9 = 9((9/10)^(n - 1) - 1). Suppose 1 - L = d for some positive number d. Then (9/10)^N - 1 > D, or (9/10)^N > E ( = d/9 + 1 ) for some constant E (E is not a function of N, only of the fixed difference d between 1 and L) for every possible choice of N. This is absurd, as it implies that after some N, (9/10)^(N + 1) either increases or remains the same as (9/10)^N which is nonsense. Thus, our assumption that there exists a positive difference (and thus any difference) between 1 and L must have been false, so the difference must be 0. \nBy the construction of real numbers, that makes 1 = 0.999... . This is a bit more tedious than proving two rational numbers equivalent (and thus why there exist many hand-waving arguments or misunderstandings of the issue), but is conceptually the same.
316406
Depends on how fast you're counting...\n\nhttp://hub1.worlded.org/docs/maththing/ny1p5.htm\nhttp://mathforum.org/library/drmath/view/58739.html
608336
L'hopital's Rule does not apply in this case as the limit can be evaluated as it is, it is not one of the special cases which you would apply that (infinity / infinity, zero / zero).  Actually, if you did apply L'hopital's Rule, you would get the wrong answer (wrong answer being 1).\n\nIn this case, since as x approaches 3 (from either side), the numerator of the function goes to zero and the denominator goes to some finite, no zero quantity, you can just plug in 3 into the function and evaluate.\n\nThe limit of the function as x --> 3 = 0
642997
The lim as x approached 7 is negative 2.\n\nThis would take my forever to type, I'll just explain. Plug in 7 and you get zero over zero. 0/0. Indeterminate form. \n\nYou can then apply L'hospital once, plug in 7 again, and you get -2.
429122
our life is based on math, but we dont notice usually, for example , when look at your points in yahoo answers and try to increase it, it means u use an exteremely mathematical logic, thank u  , bye
924358
Apparently an innocent question but otherwise a very intelligent one. Besides the arithmetic part, it has other meaning in the world of living beings which may not lead to infinity but an unknown number through propagation. Two human couples may produce any number of children up to 50 or so, insofar as the maximum is concerned. I am sure, the question is not confined to arithmetics only.
171821
I think math is wonderful, there are so many uses for it. It is an ever expanding topic too, I don't think we will ever find a "limit" to what we can do with math
226634
I don't know if this counts as an application, but it can simply be a fun exercise for a person who loves number to just sit down and do the math.  That probably seems silly to people who are more goal oriented, since they just let the computer do all the work.\n\nI love numbers, and it's more fun than just sitting around twiddling my thumbs(or doodling, since I use paper and pencil)
1056381
It's not right, it's an approximation to 3 significant digits, and for many things, that's close enough.\n\nIf you need more sigmificant digits, you can pretty much get as many as you like.
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.
1090881
Although our computing tools (such as our brain) are finite, but they can touch infinity through mathematical induction or recursion.\n\nFeeling finit ocuures for a neural network, when the data stored in it is going to overfow the nerons capacity.
888996
http://en.wikipedia.org/wiki/Indefinite_and_fictitious_large_numbers\n\nlook at that
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
1023164
Limit ( n to infinity) (n + i/n)^ n
339964
Mathematical Analysis!