25(x-3) = 25 x -75
[ 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 )
[ 25 - x^2 - 4 x - 4 ]/ (x-3)
[ - x^2 - 4 x + 21 ] / (x-3)
- [ x^2 + 4 x - 21 ] / (x-3)
- (x-3)(x+7) / (x-3)
-x - 7
-3 - 7
PLEASE US PARENTHESES SO WE CAN TELL NUMERATOR FROM DENOMINATOR !!!!!
sorry! the question was
lim [25 - (x+2)^2] / [x-3]
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.
Calculus - Evaluate each limit. If it exists. a) Lim x^3 + 1 x->1 ------- x...
math - i need some serious help with limits in pre-calc. here are a few ...
Calculus. Limits. Check my answers, please! :) - 4. lim (tanx)= x->pi/3 -(...
Calculus - Evaluate the following limit. lim e^(tanx) as x approaches the ...
Calculus, please check my answers! - 1. Evaluate: lim x->infinity(x^4-7x+9)/(...
Calculus help, please! - 1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0...
Calculus - Evaluate the limit: lim ((7-x)^2)-9 / 4-x x->4
calculus - Evaluate the limit: lim as x -> -2 (x^4 - 16) / (x+2)
Calculus - evaluate the limit: lim cos(x + pi/2)/x x->0
Calculus - Evaluate the limit. lim x-->infinity 5x+6/6x^2-3x+7