Question

calculate the limit in the following exercise, using a table. verify your answer by using a graphing calculator.
lim(x→3) (x² + 4x - 21)/(x² - 9)
let f(x)=(x² + 4x - 21)/(x² - 9). complete the table below.
x 2.9 2.99 2.999 3.001 3.01 3.1
f(x) blank blank blank blank blank blank
(round to three decimal places as needed.)

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+4x - 21=(x + 7)(x-3)$ and the denominator $x^{2}-9=(x + 3)(x - 3)$. So $f(x)=\frac{(x + 7)(x - 3)}{(x + 3)(x - 3)}=\frac{x+7}{x + 3},x
eq3$.

Step2: Calculate $f(x)$ for each $x$ - value

For $x = 2.9$:
$f(2.9)=\frac{2.9+7}{2.9+3}=\frac{9.9}{5.9}\approx1.678$.
For $x = 2.99$:
$f(2.99)=\frac{2.99+7}{2.99+3}=\frac{9.99}{5.99}\approx1.668$.
For $x = 2.999$:
$f(2.999)=\frac{2.999+7}{2.999+3}=\frac{9.999}{5.999}\approx1.666$.
For $x = 3.001$:
$f(3.001)=\frac{3.001+7}{3.001+3}=\frac{10.001}{6.001}\approx1.667$.
For $x = 3.01$:
$f(3.01)=\frac{3.01+7}{3.01+3}=\frac{10.01}{6.01}\approx1.666$.
For $x = 3.1$:
$f(3.1)=\frac{3.1+7}{3.1+3}=\frac{10.1}{6.1}\approx1.656$.

Answer:

x	f(x)
2.99	1.668
2.999	1.666
3.001	1.667
3.01	1.666
3.1	1.656

The limit $\lim_{x
ightarrow3}\frac{x^{2}+4x - 21}{x^{2}-9}=\frac{3 + 7}{3+3}=\frac{10}{6}=\frac{5}{3}\approx1.667$ (verified by the values in the table).