calculus

posted by .

Recall that a function G(x) has the limit L as x tends to infinity, written
lim(x->infinity)G(x) = L,

if for any epsilon > 0, there exists M >0 so that if x > M, then |G(x) − L| < epsilon.

This means that the limit of G(x) as x tends to infinity does not exist if for
any L and positive M, there exists epsilon > 0 so that for some x > M,
|G(x) − L| >(or equal to) epsilon.

Using this definition, prove that
the indefinite integral of sin(theta)
diverges.

[Hint: Consider the cases L >(or equal t0) 1 and L < 1 in order to deal
with all possible L values.]

  • calculus -

    The integral of sinx is from 2pi to infinity

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus - ratio test

    infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
  2. calculus - ratio test

    infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) …
  3. Pre-cal

    Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity lim 4n-3 / 3n^2+2 n-> infinity I did lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity (2+6x-3x²)/(4x²+4x+1) …
  4. Calc. Limits

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
  5. calc

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity?
  6. Calculus

    Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
  7. calculus

    Recall that a function G(x) has the limit L as x tends to infinity, written lim(x->infinity)G(x) = L, if for any epsilon > 0, there exists M >0 so that if x > M, then |G(x) − L| < epsilon. This means that the limit …
  8. calculus

    State which of the conditions are applicable to the graph of y = f(x). (Select all that apply.) lim x→infinity f(x) = −infinity lim x→a+ f(x) = L lim x→infinity f(x) = L f is continuous on [0, a] lim x→infinity …
  9. Check my CALCULUS work, please! :)

    Question 1. lim h->0(sqrt 49+h-7)/h = 14 1/14*** 0 7 -1/7 Question 2. lim x->infinity(12+x-3x^2)/(x^2-4)= -3*** -2 0 2 3 Question 3. lim x->infinity (5x^3+x^7)/(e^x)= infinity*** 0 -1 3 Question 4. Given that: x 6.8 6.9 6.99 …
  10. Math

    Find the limit if it exists. lim 1/(x-2) = infinity x→2+ lim 1/(x-2) = negative infinity x→2- lim 1/(x-2) = Does not exist x→2 lim (3x+2) = infinity x→∞ lim 999/(x^3) = 0 x→-∞ Can someone check …

More Similar Questions