Calculus

posted by .

Evaluate the limit. as x approaches infinity, lim sqrt(x^2+6x+3)-x

  • Calculus -

    sqrt(x^2+6x+3) =

    x sqrt(1 + 6/x + 3/x^2)

    Using the series expansion:

    (1+y)^p = 1 + p y + p (p-1)/2 y^2 + ...

    for p = 1/2 gives:

    sqrt(1 + 6/x + 3/x^2) = 1 + 3/x +

    O(1/x^2)

    Where the O(1/x^2) means that there exists an R and a constant c such that for x larger than R, the difference between the square root and (1+3/x) becomes less than c/x^2.

    We then have:

    x sqrt(1 + 6/x + 3/x^2) - x =

    3 + O(1/x)

    This means that the limit for x to infinity is 3.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    Note that pi lim arctan(x ) = ---- x -> +oo 2 Now evaluate / pi \ lim |arctan(x ) - -----| x x -> +oo \ 2 / I'm not exactly sure how to attempt it. I have tried h'opital's rule but I don't believe you can use it here. Any help …
  2. Calculus

    Evaluate the following limits. lim as x approaches infinity 6/e^x + 7=____?
  3. Calculus

    Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0... Should i use the formula e= lim as x->0 (1+x)^(1/x) or e= lim as x->infinity (1+1/n)^n Am i able to substitute in x/n for x?
  4. Calculus

    Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0... Should i use the formula e= lim as x->0 (1+x)^(1/x) or e= lim as x->infinity (1+1/n)^n Am i able to substitute in x/n for x?
  5. calculus

    evaluate lim of x as it approaches infinity for sqrt of x^2 -1/2x+1
  6. AP Calculus

    if i define the function f(x)= x^3-x^2-3x-1 and h(x) = f(x)/g(x), then evaluate the limit (3h(x)+f(x)-2g(x), assuming you know the following things about h(x): h is continuous everywhere except when x = -1 lim as x approaches infinity …
  7. calculus

    if i define the function f(x)= x^3-x^2-3x-1 and h(x) = f(x)/g(x), then evaluate the limit (3h(x)+f(x)-2g(x), assuming you know the following things about h(x): h is continuous everywhere except when x = -1 lim as x approaches infinity …
  8. Calculus help, please!

    1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 9 The limit …
  9. Calculus, please check my answers!

    1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 …
  10. 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 …

More Similar Questions