If p is a polynomial show that lim x→ap(x)=p(a

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

Solution. Notice that

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

  x = 1 and x = 4

Step-by-step explanation:

A graph of the given equation, after rewriting to make the solutions be zeros of the function, shows the solutions to be x = 1 and x = 4.

You can take the antilog and solve the quadratic:

  log3((6x)^2/(4x)) = log3((x+2)^2) . . . . consolidate the logs

  9x = (x +2)^2 . . . . simplify and take the antilog

  0 = x^2 -5x +4 . . . subtract 9x

  0 = (x -4)(x -1) . . . . factor

Solutions are x = 1 and x = 4. (These values make the factors zero.)