Prove this restriction?
913663
Let a = 3, b - 5, and n = 3.  Then a^n + b^n = 3^3 + 5^3 = 27 + 125 = 152 = 441 - 289 = 21^2 - 17^2, which are coprime squares.  So maybe you left out something?  Check your question.
910126
I'm assuming this is with metric spaces D with the metric d and ℝ with the Euclidean metric.\n\nProof.\n\nLet ε > 0. Since both f and g are uniformly continuous, ∃ δ₁, δ₂ > 0 such that for ∀ x, y ∈ D,\n\nd(x, y) < δ₁ ⇒ |f(x) - f(y)| < ε / 2, and\nd(x, y) < δ₂ ⇒ |g(x) - g(y)| < ε / 2.\n\nLet δ = min(δ₁, δ₂). Then d(x, y) < δ implies\n\n|(f+g)(x) - (f+g)(y)|\n= |f(x) + g(x) - f(y) + g(y)|\n= |f(x) - f(y) + g(x) - g(y)|\n≤ |f(x) - f(y)| + |g(x) - g(y)|\n< ε / 2 + ε / 2\n= ε. ∎
166307
first prove that they do\nthen i will prove that wrong
838624
Finite but unbounded.
954661
Suppose f : D → R has a limit at x0. Prove that | f | : D → R has a limit at x0 and that limx→x0 | f(x) | = | limx→x0 f(x) |.\n\n\nLet limx→x0 f(x) = c\n\nNow | f(x) | = f(x) for f(x) ≥ 0 = c\n= -f(x) for f(x) < 0 = -c\n\nSo limx→x0 | f(x) |  \n\n= limx→x0 f(x) for f(x) ≥ 0 = c\nlimx→x0 -f(x) for f(x) <0 = - limx→x0 f(x) x < 0 = -c\n\n= |c|\n\n= | limx→x0 f(x) | .................... QED
691620
(sigh)... Req = R1 + R2 + ...\n\nReq = 5 + 3 = 8.\n\nYay, I win.
910422
no \n3x-1>1 \n3x>2 \nx>2/3
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
1081259
Limt { [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nMultiple by 9/9\n\nLimt {9/9 [ sin^2 (3t) ] / [t^2 }\nt-->0\n\nLimt {9 [ sin^2 (3t) ] / [9t^2 }\nt-->0\n\n9 Limt { [ sin^2 (3t) ] / [3t]^2 }\nt-->0\n\nsince\nLimt  sin u /u = 1\nu-->0\n\nThen let 3t = u and as t-->0 so u-->0 \n\n9 Limt { [ sin (u) ] / [u] }^2 \nu-->0\n\n9 Limt { [ sin (u) ] / [u] }* { [ sin (u) ] / [u] }  \nu-->0\n\n9 *1*1 = 9
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
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)
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
271648
boy we carry those text books along with us\nif u want us to prove then provide the theorem plz
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0