evaluate the limit x ->0 of  (sin x)^x?
454694
1\nDetails:\nLet y = lim x->0 of (sinx)^x\nTake the natural log of each side.\nln(y) = lim xln(sinx)\nWrite xln(sinx) as (ln(sinx))/(1/x)\nas x goes to 0, this goes to infinity over infinity.\nUse L'Hopital's rule \n(ln(sinx))/(1/x) becomes\n(cosx/sinx)/(-x^(-2))\n= (-x^2)(cosx)/(sinx)\nAs x goes to 0, this goes to 0/0.\nUse L'Hopital's rule  again\n(-x^2)(cosx)/(sinx) becomes\n[(x^2)(sinx) - 2x(cosx)]/(cosx)\nPlug in 0.\n[0-0]/1 = 0\nThis equals ln(y)\nln(y) = 0 means y = 1
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"
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 → ∞ )
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
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
892981
since this limit is in a indeterminant form of 0/0 you can apply L'hospital's (sp?) rule\nwhich is taking the derivitive of the top and the derivitive of the bottom and evaluating the limit\n\nso this becomes [1/(1+x)]/1\nwhich is Lim x->0 1/(1+x) \n= 1/1 = 1\n\nedit:\nbah in that case im not sure...\nmaybe try a table of values and show that from each side of zero the limit approaches 1
1116129
As x tends to 0 [1/x] tends to infinity. So 2^[1/x] is 2^[infinity](almost). So the limit naturally becomes infinitely large.
910686
lim x →∞ [x ^ ln x / ln x ^ x]\n= lim x →∞ [x^ ln x / xln x]\n= lim x →∞ [x^ (ln x - 1) / ln x]\n= lim x →∞ [e^{ln x * (ln x - 1)} / ln x]\n= lim x →∞ [{1/x(ln x - 1) + ln x*1/x}*e^{ln x * (ln x - 1)}/1/x] by l'Hopital's Rule\n= lim x →∞ [(2ln x - 1) * e^{ln x * (ln x - 1)}\n= lim x →∞ [(2ln x - 1) * x^ln (x/e)]\n= ∞\n\nPlease ignore as according to Excel this is clearly incorrect!!\nThe limit = 0\n\nMy error arose from interpreting lnx ^ x as ln(x^x) rather than (ln x) ^ x\n\nLet x = e^p. Then lnx = p\nlim x →∞ [x ^ ln x / ln x ^ x] Let x = e^p Then lnx = p\n= lim p →∞ [e^p²/p^(e^p)] and as x→∞, p→∞\n= lim p →∞ [e^p²/e^(lnp*e^p)]\n= lim p →∞ [e^(p² - lnp.e^p)]\nExamining the index p² - lnp.e^p\nLim p →∞ (p² - lnp.e^p)\n= Lim p →∞ (2p - 1/p.e^p - lnp.e^p)\n= -∞\nSo lim p →∞ [e^(p² - lnp.e^p)] = 0
790936
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
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^_^
613021
http://www.ask.com/web?q=limit%28%28%28n%21%29%5E%281%2Fn%29%29%2Fn%29%2Cn%3Dinfini...+&qsrc=0&o=0
914435
x > 5\n\n- - - - - -s-
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.
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0
6900
-infinite < x < +infinite\n-1 < sin(x+2) < 1
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.