Analysis ? involving limits of functions and sequences?
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Suppose f : D → R has a limit at x0. Prove that | f | : D → R has a limit at x0 and that limx→x0 | f(x) | = | limx→x0 f(x) |.\n\n\nLet limx→x0 f(x) = c\n\nNow | f(x) | = f(x) for f(x) ≥ 0 = c\n= -f(x) for f(x) < 0 = -c\n\nSo limx→x0 | f(x) |  \n\n= limx→x0 f(x) for f(x) ≥ 0 = c\nlimx→x0 -f(x) for f(x) <0 = - limx→x0 f(x) x < 0 = -c\n\n= |c|\n\n= | limx→x0 f(x) | .................... QED
711149
Is it lim (x approaches 0)?
1081259
Limt { [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nMultiple by 9/9\n\nLimt {9/9 [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nLimt {9 [ sin^2 (3t) ] / [9t^2 }\nt-->0\n\n9 Limt { [ sin^2 (3t) ] / [3t]^2 }\nt-->0\n\nsince\nLimt  sin u /u = 1\nu-->0\n\nThen let 3t = u and as t-->0 so u-->0 \n\n9 Limt { [ sin (u) ] / [u] }^2 \nu-->0\n\n9 Limt { [ sin (u) ] / [u] }* { [ sin (u) ] / [u] }  \nu-->0\n\n9 *1*1 = 9
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Yes, linear functions have limits.\n\nFor example, the limit of 3x+2 as x approaches 4 is 14. Here's how to prove it:\n\nFor every e>0, choose d=e/3. Then, if 0<|x-4|<d, we have\n|(3x+2)-14|=|3x-12|=3|x-4|\n<3d=e.\n\nFor this situation, we say that the limit of a function f(x) as x p&#92;approaches a is L is for every tolerance e>0, there is a tolerance d>0 so that whenever 0<|x-a|<d, we have\n|f(x)-L|<e.
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Since f ist continous on the compact set [0,1] it must be bounded by some value M. The integral over x^n*|f(x)| is then smaller than the integral over M*x^n, which is M/(n+1) and goes to 0.\nThus, your original series must also go to 0 when n tends infinity.
203992
a) [1,2] is a sub set of domain of g so for every x in [1,2]\nwe have a unique value of g.\n\nb) if the sequence converge, in the limit Xn+1 must be equal to X, so if the limit is L we have:\n\nL=(3L^2+3)^1/4\n\nL^4=3L^2+3\n\nL^4-3L^2-3=0\n\ndelta=9+12=21\n\nL^2=(3+√21)/2  (L^2>0)\n\nL = √[(3+√21)/2]  (L>0 because X0>0)\n\n4<√21<5  ---> 7<3+√21<8\n\n---> 3<(3+√21)/2<4  ---> 1<L<2\n\nc) we have (Xn+1)^4=3Xn^2+3 let Yn be Xn^2\n\n---> (Yn+1)^2=3Yn+3 ---> Yn
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hav u ever heard of the Principle of Mthematical Induction ?\nthat could solve ur problem in a jiffy!!\n\nHowever if u havn't, dont worry!!\n\nNote that, the sum in ur series(i.e. 1+1/2+1/3+...+1/n),where n>=2, is always less than 2, but greater than 1. so it can't b an integer.\nwanna know why?\ntry cramming up some theorems about Sequence & series of real numbers!! its a regulation part of any under-graduate level Real analysis course.
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You can construct one simply by interleaving a bounded sequence with an unbounded sequence. For example, the sequence 1/n with the sequence n to form \n\nS(i) = if i is odd 1/i if i is even i. \n\nThis has 1/i as a convergent subsequence.
274258
An integral with which the limits are known, as oppossed to an indefinite integral.
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
219346
The first link below is a problem discussing improper integrals (which classifies all integrals with an infinite limit of integration). Just scroll to the bottom of the page--the top of the page is a different problem altogether. Basically, the tip is to replace the "troublesome" limit values with a constant (like k) and then evaluate the limit as k approaches that value. In this case, k becomes one of the limits of integration and you evaluate the integral as k approaches infinity.\n\nI recommend The Complete Idiot's Guide to Calculus (follow the second link in the Sources list below) when brushing off calculus rust, and not only because I am the author of the book. :) It's a light-hearted but content-heavy review of calculus, beginning with advanced algebra review and ending with function approximation by infinite series approximation.\n\nAlso, spend some time with "Tutorials for the Calculus Phobe," free Flash-based calculus tutorials on my website, Calculus-Help.com. They're fun but help you conceptualize calculus basics. The third and final link in the sources houses the Phobe tutorials.\n\nBest of luck!
589983
Add some rigor. Remember that a sequence {a_i} is defind to converge to a limit L iff for every positive difference D > 0, there exists some n such that for all N > n, the difference between the limit L and any subsequent term a_N is less than D. We are squeezing D down to 0 by showing that we can always supply such an n. Thus, your job is to supply some pseudofunction n(D) giving an upper range of indices that satisfy the definition above.\n\nFor the pattern above, the (2n + 1)th term is 8 + 1/(n + 2) while the (2n)th term is 8 + 2/(n + 2). As n increases without bound, the (2n + 1) subsequence approaches 8, as given any positive difference D, one can find a least (2n + 1) such that all terms greater than (2n + 1) yields a difference 8 - (8 + 1/(n + 2)) = 1/(n + 2) less than D. In this case, choose n(D) > (1 - 2D)/D. \nUse a similar tactic to show that the even indexed subsequence converges to the same value, and thus the sequence converges to 8.\n\nAs for the second sequence, as long as all of the remaining values are 6, it converges to 6 by a trivial argument. Given a positive difference D, simply choose n(D) > 6, since 6 - a_n = 0 for any n > 6.
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http://www.ask.com/web?q=limit%28%28%28n%21%29%5E%281%2Fn%29%29%2Fn%29%2Cn%3Dinfini...+&qsrc=0&o=0
891612
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.
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I guess you probably use the Epsilon definition for convergence? Then, you know:\nFor every Epsilon greater than 0, there exists an N such that n>N implies /xn - a/ < E'.\nYou want to show that this also implies:\n/xn^(1/3) - a^(1/3)/ < E.\n\nI think that you can probably do something with:\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\nmaking x = a in this equation, the proof's a = b.\nThen, you can find an N so that /[xn^(1/3) - a^(1/3)][xn^2/3 + xn^(1/3)a^(1/3) + a^(2/3)]/ < E'. Perhaps this is easier to relate to the first one?\n\nGood luck!
348048
A sequence {x_n} is Cauchy if given e>0 we can find an integer N such that || x_m - x_n || < e for all m,n > N.  I want to prove a lemma first:\n\nLEMMA: Let f: A -> R^m be uniformly continuous.  If {x_n} is a Cauchy sequence, then so is { f(x_n) }.\n\nProof: Let e > 0 be given.  We must show that there exists an integer N such that || f(x_m) - f(x_n) || < e whenever m, n > N.  Since f is uniformly continuous, we can find d > 0 such that || f(x) - f(y) || < e whenever || x -y || < d.  Since d is given -- and only depends on e! -- there is an integer N such that || x_m - x_m || < d for all m, n > N.  This means || f(x_m) - f(x_n) || < e for all m, n > N.\n\nWe now prove the main problem with the following.\n\nPROPOSITION: A sequence {x_n} is Cauchy if and only if lim x_n exists.\n\nProof: First assume that L = lim x_n exists.  Given e > 0, there exists an integer N such that || x_n - L || < e/2 whenever n > N.  For m, n > N, we have, by the Triangle Inequality,\n\n|| x_m - x_n || <= || x_m - L || + || x_n - L || < (e/2) + (e/2) = e.\n\nConversely, say that {x_n} is a Cauchy sequence.  We give a Proof by Contradiction, so assume that lim x_n does *not* exist.  That means given any point L in R^m, real e > 0, and integer N we can find an integer m > N such that || x_m - L || > e.  Since e is given, we can find an integer N such that || x_m - x_n || < e for all m, n > N.  Choose n = N+1 and L = x_n.  Then there exists an integer m > N such that\ne < || x_m - L || = || x_m - x_n || < e\n\nwhich is a contradiction.  Hence lim x_n must exist.\n\nCOROLLARY: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then lim f(x_n) exists.\n\nProof: If {x_n} is Cauchy and f: A -> R^m is uniformly continuous, then { f(x_n) } is also Cauchy.  The limit of a Cauchy sequence exists, so L = lim f(x_n) exists.
807703
Achieved Theoritical Upper Limit