lim x -> 0 of e^x + (e^-x) - x^2 - 2 divided by sin^2 x-x^2?
404485
You can solve this essentially using L'Hospital's Rule.\n\nDenote the numerator by\n\n   f(x) = e^x + e^{-x} - x^2 - 2 = (1/12) x^4 + (1/360) x^6 + ...\n\nI found this using the Taylor series expansion e^x = 1 + x + (1/2) x^2 + (1/6) x^3 + (1/24) x^4 + (1/120) x^5 + ...  Denote the denominator by\n\n   g(x) = (sin x)^2 - x^2 = (-1/3) x^4 + (2/45) x^6 + ...\n\nI found this using the Taylor series expansion sin x = x + (-1/6) x^3 + (1/120) x^5 + ...  Hence we find the ratio\n\n   f(x) / g(x) = ( (1/12) +  (1/360) x^2 + ... ) / ( (-1/3) + (2/45) x^2 + ...)\n\nwhere the powers x^4 cancel out.  As x ->0, the ratio f(x) / g(x) doesn't tend to f(0) / g(0) because f(0) = g(0) = 0.  However, the ratio f(x) / g(x) is still defined at x = 0, so we see that f(x) / g(x) tends to (1/12) / (-1/3) = -1/4.
944032
Did you make a typo? 1 - 1 = 0. Maybe you mean lim x --> 0 of \n\n     ( 1 - 1/x ) sin x cos( x/2 )\n\n     = sin x cos (x/2) - ( sin x / x ) cos( x/2 ) \n\nLimits as x --> 0 of sin x, cos( x/2 ) and sin x / x are 0, 1 and 1, so answer is\n\n     0 × 1 - 1 × 1 = -1
1057515
Please check the statement of the problem, then apply this.\n\nWe know that\nsin (x)=(e^(ix)- e^(-ix))/2i\n\ncos (x)=(e^(ix)+ e^(-ix))/2\n\nalso\n\nsin(ix)=i(sinh(x))=(e^(-x)- e^(ix))/2i\n\ncos(ix)=(cosh(x))=(e^(-x)- e^(ix))/2\n\nThe only thing you need is if z=x+iy then |z|=sqrt(x^2 + y^2)\n\nGood luck
755844
1 correct \n\n2 ∫x/e^x dx \n= ∫x*e(-x) dx\n= -xe^(-x) - ∫-e^(-x). 1.dx\n= -xe^(-x) - e^(-x) + c\n\nSo∫[x = 1 to ∞] x/e^x dx = lim t →∞ (-te^(-t) - e^(-t)) - (-1/e - 1/e)\n= 2/e (NB lim t →∞ (-te^(-t)) = o by d'Hopitals rule)\n\n3 correct\n\n4. Use (n² + 1)/[e^n * (n + 1)²] \n=  [(n² + 1)² - 2n]/[e^n * (n + 1)²]\n= 1/e^n - 2n/e^n(n + 1)² \nNow 1/e^n < 1/3^n and  2n/e^n(n + 1)² < 2n/3^n*n² \n= 2/n*3^n \nSo both parts converge so the difference converges\n\nNOTE: It is many, many years since I have done these BUT I am confident that I am correct ... Maybe you can tell me if that is so!
867883
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
711149
Is it lim (x approaches 0)?
749046
well the term you've quoted is a constant...\n\nso (e^2/1+e^2)x would be the answer to that question.\n\nbut i'm guessing you typed it wrong, or are confused.
953356
Let f(x) = sin (1/x) and g(x)=-sin (1/x). Neither f nor g have limits at 0, but [x→0]lim f(x) + g(x) is obviously 0.
518111
As x->0, -3x->0. \nNow let -3x = z.\nThus the limit becomes\nLim z->0 [-3*{ln(1+z)}/z]\n= (-3)*Lim z->0 [{ln(1+z)}/z]\n= (-3)*1\n= -3\n\nSo 'c' is correct.
798431
limit x>0 [(3x)sin3x/3x*(4x)sin4x/4x]\n/[(6x)tan6x/6x(x)sinx/x]\nlimit x>0 3x*4x/6x*x\nlimit x>0 12x^2/6x^2\n=2 (answer)
896031
Why not?  If you divide top and bottom of this expression by x^7 you get\n(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))\n\nEvery term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,\ntherefore it approaches 3/9.
951210
ok...well 1/x becomes basically infinity as x gets really close to zero...so (1/x)/x woud be infinity/something really close to 0 which would make the value infinity\n\nso we have\n\ne-(1-x)^infinity\nwell...thats ok\n\ne-(1 - (basically zero))^infinity is\ne-1^infinity\nwhich is\ne-1\n\nso the limit as x approachs zero from the right is e-1 which is approximately 1.71828\n\nhope this helps\nmatttlocke
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.
1048224
This question is really only interesting if what you MEANT to type was:\n(4 - x^2) / (x + 2)\n\nSo you want to solve:\n\n. lim (4 - x^2) / (x + 2)\nx→ -2\n\nOf course the original expression is not defined at x = -2, because that would involve dividing by zero. However, for any real numbers x ≠ -2:\n(4 - x^2) / (x + 2) = (2 - x)(2 + x) / (x + 2)\n\nSince x ≠ -2 it is valid to divide top and bottom of the fraction by the common term (x + 2) which leaves:\n\n(4 - x^2) / (x + 2) = (2 - x), for all x ≠ -2.\n\nThus,\n. lim (4 - x^2)/(x + 2)\nx→ -2\n= lim (2 - x) = 2 - (-2) = 4 <<<<<<<<<<<\n. x→ -2
708189
the integral diverges,\nsince lim x->+-infinity e^(pix^2)  = infinity\ns
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.
1060628
simple find the integral of pi*(e^(-x^2))^2 from a=0 and b=2 since as x goes to infiniety is goes to zero but it is very close to zero when x=2 you can find this by graphing f(x) \nanswer should be 0.9188 good luck!