limit as x tends to 0 of incosx/x^2?
346514
Do you mean the limit of ln(cos x)/(x^2) as x goes to 0?\n\nIf so, substituting 0 in for x gives 0/0, which means you can use L'Hopital's rule.\n\nTo do so, take the derivative of the numerator and denominator, which are respectively -sinx/cosx=-tanx and 2x.\n\nThat still gives 0/0, so we can apply L'Hopital's rule again.  This time, we get -1/(cosx)^2 for the derivative of the numerator, and 2 for the derivative of the denominator.\n\nPlugging x=0 into -1/(cosx)^2 / 2 gives -1/2.\n\nSo the limit of ln(cos x)/(x^2) as x goes to 0 is -1/2.
962773
Use L'Hospital's rule twice and get e^x/2 which has a limit of infinity.
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.
1051867
The range is:\n(-1)<x<=(1)\nx=1 is admissible
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
497029
No, because the limit as x->2- is undefined.  The limit exists for x->2+.
879179
There is no limit.\n\nYou can see this by approaching from two different directions, along the x-axis and along the y-axis.If x is zero and y->0, then the formula becomes:\n\n   (siny)^2/y^2\n\nSince sin y/y -> 1 as y->0, this also approaches 1 as y approaches zero.\n\nOn the other hand, if y=0, and x->0, then the formula becomes:\n\n    x^2/(2x^2)\n\nwhich has a limit of 1/2 as x->0.
477866
Let y=cos2x, then, \ndy/dx = limit (as h tends to zero) [(cos2(x+h) - cos2x)/h]\n      = lim. [(cos2x cos2h  - sin2x sin2h - cos2x)/h]\n      = lim. [(cos2x(cos2h-1) -sin2x sin2h)/h]\n      = lim. [(cos2x (-2(sin2h)^2)/h] - sin2x* lim.(sin2h/h)\n      = -2 cos2x * (lim. sin2h/h)*(lim. (sin2h))- 2 sin2x\n      = -2 cos2x * 2 * 0 - 2 sin2x\n      = -2 sin2x
807227
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 → ∞ )
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.
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^_^
778101
x^x = e^{xln(x)} \nnow, the lim x->0 x ln(x) = 0,\nso lim x->0 x^x = e^0 =1\n`\n`
712410
limit theta tending to 0 sintheta/theta =1
147502
log x goes to negative infinity\nbut with the squared it becomes positive.\nSo the limit is +infinity
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.
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
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.