if 'a' does not equal 0, then limit as x approaches a of (x^2 - a^2) / (x^4 - a^4) is what?
830308
Denominator can be written as (x^2-a^2).(x^2+a^2)\n\nThe x^2-a^2 terms cancels when a is not equal to zero leaving the formula as\n\n1/(x^2+a^2)\n\nThe denominator here approaches zero as x approaches a \nso the original formula tends to infinity.  Note that since squares are positive the answer is positive infinity as x approaches a from the left and the right.\n\nOther ways are to use L'Hopitals rule.
1023164
Limit ( n to infinity) (n + i/n)^ n
984623
It will be 1/(x^2)\n\nFor example\n\n(1/2)^2 = (1^2)/(2^2) = 1/ 4\n(1/3)^2 = (1^2)/(3^2) = 1/ 9\n(1/4)^2 = (1^2)/(4^2) = 1/16\n\nSo the result actually gets smaller even though you are squaring it
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
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.
914503
You should remember how to do this in the future because the limit of this expression as h approaches 0 gives you the derivative or slope of f(x) at x = 4.\n\n[ f(4 + h) - f(4) ] / [ (4 + h) - 4 ]\n\n[ (4 + h)^2 - 4^2 ] / h\n\n(h^2 + 8h + 16 - 16) / h\n\nh + 8
793364
To show that it doesn't exist, look at the path where x=0 and y moves to zero.  The problem reduces to:\n\nsin(y)/y^2\n\nYou can use L'Hospital's rule to show that this limit does not exist.
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
914435
x > 5\n\n- - - - - -s-
66783
(x-2)^2=x\nor x^2 - 4x + 4=x\nor x^2 - 5x + 4=0\nor x^2 - 4x - x + 4=0\nor x(x-4) - 1(x-4)=0\nor (x-4)(x-1)=0\n.'. x=4 or x=1;
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.
22472
It's well known that lim x->0 sin(x)/x=1\nThen lim x->inf. sin(1/x)/(1/x)=1\nwhich is exactly what you need.
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"
490253
infinity/0 is infinity.  If you have any function f(x) and g(x) where f(x) goes to infinity as x -> k and g(x) goes to zero as x -> k, then 1/g(x) goes to infinity as x -> k, so\n\nf(x)*(1/g(x)) -> infinity as x -> k.\n\ninfinity*zero depends. If you look at 1/x and x^2, then as x -> 0, 1/x -> infinity and x^2 -> 0, but\n\n(1/x)*(x^2) = x, so its limit is 0 as x -> 0.\n\nIf you look at 1/x^2 and x, then as x -> 0, 1/x^2 -> infinity and x -> 0, but\n\n(1/x^2)*(x) = 1/x, so its limit is infinity as x -> 0.
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
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^_^