what is the limit of x as x tends to positive infinity of  [3^x-3^-x]/[3^x+3^-x]?
409697
As x becomes very large, 3^(-x) becomes very small compared to 3^x, and can thus be neglected. Therefore, you have (3^x)/(3^x) = 1
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.
784830
just another math guy may be correct, but there is a much simpler solution using l'Hospital's rule.\n(In this proof, all limits are as n→∞)\nf(n) = (1+1/n)^n = [(n+1)/n] ^ n\n= (n+1)^n / n^n\nTake the derivitive of numerator and denomerator to find an equivalent limit.\nlim f(n) = lim [(n+1)^n * (n/(n+1) + ln(n+1))] / [n^n * (1+ln(n))]\n= lim (n+1)^n / n^n * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }\n= lim f(n) * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }\n\nTherefore, lim { [n/(n+1) + ln(n+1)] / [1+ln(n) ] } = 1 !\nor lim n/(n+1) + ln(n+1) = lim (1 + ln(n) )\nor, in the limit, \nln(n+1) - ln(n) = 1/(n+1)\nln[(n+1)/n] = 1/(n+1)\ne^(1/(n+1)) = (n+1)/n = 1+1/n\nor e = lim (1+1/n)^(n+1) = lim (1+1/n)^n
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
679340
By definition, you would do the following:\n\nf'(x) = lim(h->0) (f(x+h) - f(x))/h\nf'(x) = lim(h->0) ((3(x+h) - 3x)/h\nf'(x) = lim(h->0) 3h/h\nf'(x) = lim(h->0) 3\nf'(x) = 3\n\nOr, you could simply do use the following fomulas:\n\nd/dx[af(x)] = a(d/dx[f(x)])\nd/dx[x^n] = nx^(n-1)\n\nd/dx[3x] = 3(d/dx[x]) = 3(d/dx[x^1]) = 3*(1*x^0) = 3\nf'(x) = 3
1005731
Here ya go buddy.  As x tends to infinity, only the highest power terms will dominate.  All you have to do is compare the highest power terms in under each root.  Effectively, the highest power of x in the first term is x^1 (since sqrt(x^2)=x) and the highest power in the second term is sqrt(x).\n\nNow, as x->inf, you are effectively looking at the function x-sqrt(x).  Since there are no denominator terms, this function goes straight to infinity.\n\nAlso, you could multiply by the conjugate to get \n(x^2+7x+50)/((sqrt(x^2+8x+50)+sqrt(x+2))\nbut you'd still have to go through the same kind of rationale, where you would be dealing with:\nlim x->inf  x^2/x = x \nwhich still goes to infinity.  Only use the conjugate if you can't make this kind of connection or if you are evaluating your limit at a finite value.
945609
Use l'Hospital's Rule when dealing with 0/0 or intfy/infty: in such cases,\n\nlim (f(x)/g(x)) = lim (f'(x)/g'(x)) if the latter exists (you may have to repeat this step using the second derivatives f''/g'').\n\nIn your first example, as x --> 1, \n\nlim (x+ x^2 + x^3 +... + x^n - n)/(x-1) \n= lim (1+ 2x + 3x^2 +... + nx^(n-1) )/1 \n= 1+ 2 + 3 +... + n = n(n+1)/2.
942305
Okay. You have a few things that you have to consider.\n\nNegative exponents: moves it to the denominator\nFractional exponents: the denominator tells you what the root is, the numerator tells you what the root is rasied to\n\nSo: (x+1)^-2/3 rewritten would be:\n\n1/[cuberoot(x+1)^2]\n\nwhen x is seven you have\n1/[cuberoot(8)^2]\n1/[2^2]\n1/4
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.
749158
The expression you wrote has not limit as x goes to 3. Instead, it grows without bound.\n\nWhat you may have meant to write is the limit of ((x-3)/(x^2-9)).\nFor that expression, both numerator and denominator go to 0 as x goes to 3. So there is a chance that the expression will have a limit.\n\nIt turns out that it does have a limit. In fact, you can find it by first factoring the denominator (x^2 - 9) = (x+3)(x-3), then dividing numerator and denominator by (x-3) (the result is (1/(x+3)), and determining the limit of that expression as x goes to 3 (it equals 1/6).\n\nNote: The first responder (Helmut) gave you the same answer that I did (1/6) by applying l'Hospital's rule. But he did not modify the expression you gave in your question, which I said was required in order to find a limit. But he should not have applied l'Hospital's rule to a a fraction where the denominator goes to 0 at the limit, but the numerator does not. If he had recognized that the numerator has to go to 0, he likely would have done what I did, which is to guess that the numerator should be (x-3), which goes to 0 as x goes to 3. Had he done that, he would have found the same limit (1/6), since d(x-3)/dx = dx/dx = 1.
1096223
The definition of a limit is slightly different depending on whether you are looking at the limit of a function at some point&#92;, or the limit of a function at infinity, or the limit of a sequence at infinity. I will do the last.\n\nWe say that the limit of the sequence a_n as n goes to infinity is the number L if whenever e>0, there is a natrual number N so that whenever n>=N, we have |a_n -L|<e.\n\nIn other words, the limit is L if for every 'tolerance' e, we can go far enough out in the sequence (N) so that every term of the sequence farther out is within the tolerance of the purported limit.\n\nAs an example. Let a_n=1/2 +1/4 +...1/2^n.\nIt is easy to show by induction that a_n =1 - 1/2^n.\n\nGiven a tolerance e>0, choose N so that 1/e <2^N. Then, if n>=N,\nwe have |a_n -1|=1/2^n < e.\n\nThis shows that the limit L is 1. It has to be proved that limits are unique when they exist, but that is not too difficult. Any good advanced calculus course will cover this stuff.
179824
Its preety easy. The answer is (3X^2)(3^X). Just take the derivative of X^3. Don't do any thing with 3^X cuz X is an exponent and when a X is an exponent you just leave it like that.
845874
1.  This is equal to\nk*(lnx)/x, and since \n(lnx)/x --> 0, so does this expression.\n\n2. Sorry, blank on this and 3.\n\n4. (n^k)/2n = (n^(k-1))/2, and I guess as long as \nk > or = 1\nthis will approach 0.\n\nOf course, if k < 1 then it has no limit, or "approaches infinity"
252229
Any non-zero number divided by a limit approaching 0 WILL go to infinity:  n/(lim->0) is inifinity so n/0 is infinity.\nBut 0/0 is UNDEFINED.
365360
Use the ratio test. We want the absolute value of the ratio as k approaches infinity of the (k+1)th term to the kth term to be less than 1. So take the expression x^n/n^(1/5) and let n = k + 1. Stick the result in the numerator. Then let n = k and stick that result in the denominator. Now take the limit as k goes to infinity. You should just end up with x (I don't want to write the whole process out here cuz it's hard to write equations but you should get something that looks like x[k/(k+1)]^(1/5) which approaches x as k -> infinity). So if |x| < 1, you're good to go. If x is plus or minus 1, we're still not sure if the series converges or not. The x = 1 case gives us a p-series with p = 1/5, which diverges. The x = -1 case gives us an alternating series in which the magnitude of each term is always decreasing and approaches zero as n goes to infinity. Thus, it passes the alternating series test for convergence and the series converges for x = -1.\n\nSo the interval of convergence is [-1,1) and the radius of convergence is 1.
949655
cos ^2 (x) is always inferior to 1\nSo if x goes to  infinity, n=x²  we have 2^n/n\nit can be written also e^x^ln(2)/x\nAs e^x/x tends to infinity when x goes to infinity\nso it is infinity
892270
First of all, you need to actually guess what the limit will be. \nf(x) = (x+2)(x-2) / (x+2) = x-2 as long as x is not equal to -2, so a good guess would be -4, since its very close to -4 when x is very close to -2.\n\nLet ε>0. Choose δ=ε. Then if 0 < |x- -2| = |x+2| < δ, we have:\nf(x) = (x+2)(x-2)/(x+2) = x-2, since x isn't equal to -2.\nSo |f(x) - -4| = |x-2- -4| = |x+2| < δ = ε, as required.