Calculus

posted by .

Evaluate the limit:
lim 25-(x+2)^2 / x-3
x->3

  • Calculus -

    25(x-3) = 25 x -75

    so
    [ 25 x -75 - x^2 - 4 x - 4 ] /(x-3)

    [-x^2 + 21 x -79 ] / (x-3)

    (x-3) is not a factor of the numerator so I can not get rid of it in the denominator.
    as x goes to 3 the numerator becomes -25 and the denominato is 0 so undefined is the answer.
    If you mistyped it and mean
    [ 25-(x+2)^2 ]/ (x-3 )
    then
    [ 25 - x^2 - 4 x - 4 ]/ (x-3)
    that is
    [ - x^2 - 4 x + 21 ] / (x-3)
    - [ x^2 + 4 x - 21 ] / (x-3)
    - (x-3)(x+7) / (x-3)
    -x - 7
    -3 - 7
    -10
    PLEASE US PARENTHESES SO WE CAN TELL NUMERATOR FROM DENOMINATOR !!!!!

  • Calculus -

    sorry! the question was
    lim [25 - (x+2)^2] / [x-3]
    x->3

  • Calculus -

    so
    -10

  • Calculus -

    I figured you must have mistyped it because (x-3) simply had to be a factor of the numerator or the question would not have been asked.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    Evaluate the limit. lim x-->infinity 5x+6/6x^2-3x+7
  2. calculus

    Evaluate the limit: lim as x -> -2 (x^4 - 16) / (x+2)
  3. math

    i need some serious help with limits in pre-calc. here are a few questions that i really do not understand. 1. Evaluate: lim (3x^3-2x^2+5) x--> -1 2. Evaluate: lim [ln(4x+1) x-->2 3. Evaluate: lim[cos(pi x/3)] x-->2 4. Evaluate: …
  4. Calculus

    Evaluate the limit: lim ((7-x)^2)-9 / 4-x x->4
  5. Calculus

    evaluate the limit: lim cos(x + pi/2)/x x->0
  6. Calculus

    Evaluate each limit. If it exists. a) Lim x^3 + 1 x->1 ------- x + 1 b) Lim 3x^2 - x^3 x->0 ---------- x^3 + 4x^2 c) Lim 16 - x x->16 --------- (√x) - 4
  7. Calculus. Limits. Check my answers, please! :)

    4. lim (tanx)= x->pi/3 -(sqrt3) 1 (sqrt3) ***-1 The limit does not exist. 5. lim |x|= x->-2 -2 ***2 0 -1 The limit does not exist. 6. lim [[x]]= x->9/2 (Remember that [[x]] represents the greatest integer function of x.) 4 …
  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. Calculus

    Evaluate the following limit. lim e^(tanx) as x approaches the righter limit of (pi/2)

More Similar Questions