lim as x-->0 (ln(30/x)/ln(2))
125206
First, use the identities for logarithms to simplify this.  When you do, you get ln(30/x) = ln30 - ln x.  Now, divide each of those terms by ln2 and you get ln30/ln2 - lnx/ln2.  Now, take the limit as x->0.  As x goes to zero from the right (the left hand limit does not exist), ln goes to minus infinity.
493556
-1 / ( 20 ( ln( 10 + e ) * ( 10 + e )^(10 * x^2 ) ) )
8016
The statement of the problem is incomplete. In order to evaluate the limit, you need to know what value x is approaching.\n\nIf you are taking the limit as x->0 then:\nsin(x)->0, so tan(sin(x))->0, cos(x/4)->1, so the numerator goes to zero, while the denominator goes to 1 (unless the elipsis (...) indicates that there are more terms that you havend listed.) That means the the limit of the quotient will be 0.\n\nFor the second on one, [sin(x)] is the greatest integer less than sin(x) which will be -1 if x is slightly less than 0 and 0 if x is slightly greater than 0. |x-P/2| will go to P/2 so you have a limit of -4/P approaching from the left and -2/P approaching from the right. The limit doesn't exist, only the two one-sided limits.
62442
[x^2/4][2(log(x) - 1) log(x) + 1]
551630
To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368
432890
..ll<3,2>ll  = sqrt( <3,2> . <3,2> )\n= sqrt( 4(3)(3) +7(2)(2) ) \n=sqrt( 36 +28) = sqrt ( 64)\n= 8
103664
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.
565431
2x<1 , x<0.5\n -1<x\n\n     -1< x<0.5
524884
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.
280429
y=x^(2/x)\nln y = (2/x) * ln(x)\nln y = 2*ln(x)/x\n\nUse L'hospitals' rule: derivative of top divided by derivative of bottom:\n\n= (2/x)\n\nTake the limit as x--> infinity:\n\nlimit = 0\n\nTo compensate for using the natural log in the beginning, you now use this answer and put it in the form of e^x so:\n\ne^0 = 1\n\nYour answer is 1.
112000
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
57818
D.    X (greater or equal to) 4\n      2(-2)+4 = 0\nso (-infin, -2) U [4, infin)
531551
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
421844
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 → ∞ )
516430
limit x----------approaching---pi/2\n\nF(x)= 6cos(5x)sec(-9x)\n= lim x-> pi/2 6 cos(5x) / cos(-9x)\n=lim x->pi/2   -6(5) sin(5x) / 9sin(-9x)\n= -30 sin (5pi/2) /9 sin(-9pi/2)\n= -30 (1)/9(-1)\n= 10/3\n'
371720
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.
466858
lim x → 0+ (5x - tan(3x))/(x + 3sin(2x)) \n\n= lim x → 0+ [x * (5 - tan(3x)/x)/(x * (1 + 3 sin(2x)/x)]\n= lim x → 0+  [ 5 - 3* tan(3x)/3x)/ (1 + 6*sin(2x)/2x)]\n= [5 - 3][1 + 6] = 2/7\n\nlim x → 1 (x+x^2+x^3+... +x^n)/x - 1\n= lim x → 1 (1 + x + x^2 + ... + x^(n-1) - 1\n= n - 1\nIf you meant lim x → 1 (x+x^2+x^3+... +x^n)/(x - 1)\nthen lim x → 1 (x+x^2+x^3+... +x^n)/(x - 1)  → ∞\n\nlim x → 1 ((x^2) - 1) * (cot(x^2 - 1))\n= lim (x - 1) → 0 ((x^2) - 1) * (cot(x^2 - 1))\n= lim (x^2 - 1) → 0 ((x^2) - 1) / (tan(x^2 - 1))\n= 1 {Using lim u → 0 (tanu/u) = 1}\n\nlim  x → 0 (cos(2x) + cos(x) - 2)/x^2\n= lim  x → 0 (1 - 2sin^2(x) + cos(x) - 2)/x^2\n= lim  x → 0 (-2sin^2(x) + cos(x) - 1)/x^2\n= lim  x → 0 (-2sin^2(x))/x^2 + lim  x → 0 (1 - 2sin^2(x/2) - 1)/[4(x/2)^2]\n= -2 lim x → 0 (sin(x)/x)^2 - 2/4 lim  x → 0 (sin(x/2)/(x/2)^2)\n= -2 - ½\n= -2½