Limit Evaluation

Question

Lim x -> 1

x^3 + x^2 + x - 3 all over (x-1) = ?

(The answer will be an integer or whole number)

Answer 1

use a graphing calulator, desmos or TI83

look at the y value when x=1, y=6

divide x-1 into the cubic, by long hand or synthetic division to get

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

plug in x=1

1^2 +2(1) +3 = 6

or use L'Hopital's rule

take derivatives of numerator and denominator separately

then plug in x=1 to get 6

Answer 2

Direct substitution for x = 1 yields 0/0.

That means you could use L'Hopital's rule, but you can also determine if (x-1 ) is a factor of the numerator, using long polynomial division or synthetic division.

Doing the division determines that f(x) = (x-1)(x² + 2x +3)/(x-1). As x gets close to 1, but not equal to 1, the

(x-1) factors cancel out. So the limit is (1)²+2(1) +3 = 6.

Using L'Hopital's rule:

lim_x->1(x³ + x² + x -3)/(x-1) = lim_x->1(3x²+2x+1)/1 = 3(1)²+2(1)+1 = 6.

Check it out here:

desmos.com/calculator/52t0qfxqsn