What is the limit of the funtion (x^2-9)/(x-3) as x tends to 0?
98782
Just substituting x for 0 yields -9/-3 = 3.  This is not the indeterminate form 0/0 hence the limit is obtained by substitution, ie the answer is 3.
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.
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
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
665272
To build on what "Thin Kaboudit" was saying, you can take the limit of a function f(x) = 3/x as x goes to zero.\n\nLets say you decided to approach x from the positive side, slowly decreasing its value toward zero.\nThe closer and closer x gets to zero, the closer the function's value gets to [positive] infinity.\nNow lets say you decide to approach x from the negative direction, starting out with negative numbers and slow increasing them toward zero.  As you get closer and closet to zero from the negagtive side, the function goes toward negative infinity.\n\nSince the limit from the positive side does not equal the limit from the negative side (infinity does not equal negative infinity), the limit of the function does not exist since the function is not defined when x = 0.
892605
(constant)/(2^n) when n->infinity => the limit is 0(zero)\n\nPS\nNot to be a a stickler for rules but is this really laptop related ? :))
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.
824215
Did you mean -(3/4) * (x^3+2x^2-3x+1) ?\n\nLet f(x) =  -(3/4) * (x^3+2x^2-3x+1). Equivalently, f(x) = -(3/4)x^3 - (3/2)x^2 + (9/4)x - 3/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nSince f is a polynomial function (of degree 3) on x, it follows that this is continuous on the set of real numbers. Thus, by applying the limit on both sides (left and right) of x, we will obtain the same value for both sides. \n\nThat is, the left-hand Limit will be as follows: as x approaches to the left of c, we make f(x) approaches f(c) = -(3/4)c^3 - (3/2)c^2 + (9/4)c - 3/4. Similarly, the right-hand limit will be: as x approaches to the right c, we also get f(x) approaches to f(c). \n\nTherefore, both the left and right-hand limit yields to f(c); so as x approaches c in any direction, we always obtain f(c). Hence, both left and right-hand behavior of f are the same for any choice of a real valued x.
1074093
The function has zeroes at -6 and 0.  It decreases to negative infinity as x decreases from -6. It increases to infinity as x increases from zero.  The only relative maximum is at -4.
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"
974331
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
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.
748556
L'Hospita;'s rule shall not apply here as it is non zero/zero form\n\nit applies for form 0/0 or infinite.infinite\n\nlimit f(x)/g(x) when it is of above form is lim f'(x)/g'(x) \n\nfor example as x->o lim sin x/ x is of the form 0/0\n= lim x->0 d/dx(sin x)/d/dx(x)\n= cos x/ 1= cos x = 1 when x->0\n\nfor your case it is 1/0 or infinite (no l'Hospital's rule)
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
239990
There are no values of x for which the function is less than 0.  The function value is between 0 and 1 when x<=0.\n\nPlot a few values to sketch the graph: x=-2,y=1/9; x=-1,y=1/3; x=0,y=1; x=1,y=3; x=2,y=9