Question

Evaluate the function f(x) at the given numbers (correct to six decimal places). f(x) = x2 − 7x x2 − 6x − 7 , x = 0, −0.5, −0.9, −0.95, −0.99, −0.999, −2, −1.5, −1.1, −1.01, −1.001 f(0) = f(−0.5) = f(−0.9) = f(−0.95) = f(−0.99) = f(−0.999) = f(−2) = f(−1.5) = f(−1.1) = f(−1.01) = f(−1.001) = Guess the value of the limit of f(x) as x approaches −1, correct to six decimal places. (If an answer does not exist, enter DNE.) lim x→−1 x2 − 7x x2 − 6x − 7 =

Answer 1

Answer:

Step-by-step explanation:

From the given information:

We are given the function:

$f(x) = \dfrac{x^2 - 7x}{x^2 -6x-7}$

when x = 0;

$f(0 ) = \dfrac{(0)^2-7(0)}{0^2-6(0) -7}$

$f(0 ) = \dfrac{(0)}{-7}$

f(0) = 0

when x = -0.5

$f(-0.5 ) = \dfrac{(-0.5)^2-7(-0.5)}{(-0.5)^2-6(-0.5) -7}$

$f(-0.5 ) = \dfrac{0.25+3.5}{0.25+3 -7}$

$f(-0.5 ) = -1$

when x = -0.9

$f(-0.9 ) = \dfrac{(-0.9)^2-7(-0.9)}{(-0.9)^2-6(-0.9) -7}$

$f(-0.9 ) = -9$

when x = -0.95

$f(-0.95 ) = \dfrac{(-0.95)^2-7(-0.95)}{(-0.95)^2-6(-0.95) -7}$

$f(-0.95 ) = -19$

when x = -0.99

$f(-0.99 ) = \dfrac{(-0.99)^2-7(-0.99)}{(-0.99)^2-6(-0.99) -7}$

$f(-0.99 ) = -99$

when x = -0.999

$f(-0.999) = \dfrac{(-0.999)^2-7(-0.999)}{(-0.999)^2-6(-0.999) -7}$

$f(-0.999 ) = -999$

when x = -2

$f(-2) = \dfrac{(-2)^2-7(-2)}{(-2)^2-6(-2) -7}$

$f(-2 ) =2$

when x = -1.5

$f(-1.5) = \dfrac{(-1.5)^2-7(-1.5)}{(-1.5)^2-6(-1.5) -7}$

$f(-1.5 ) =3$

when x =  -1.1

$f(-1.1) = \dfrac{(-1.1)^2-7(-1.1)}{(-1.1)^2-6(-1.1) -7}$

$f(-1.1 ) =11$

when x = -1.01

$f(-1.01) = \dfrac{(-1.01)^2-7(-1.01)}{(-1.01)^2-6(-1.01) -7}$

$f(-1.01 ) =101$

when x = -1.001

$f(-1.001) = \dfrac{(-1.001)^2-7(-1.001)}{(-1.001)^2-6(-1.001) -7}$

$f(-1.01 ) =1001$

So, if we make a table to show the correlation for the values of  x and corresponding f(x) and also include the value of -1 whose limit of f(x) is unknown, we have;

x          0       -0.5       -0.9       -0.95        -0.99         -0.999       -1

f(x)       0       -1            -9        -19               -99               -999       ????

x         -1.001        -1.01              -1.1              -2

f(x)       1001           101               11                 2

From the table above, we can see that for x values ranging from 0 → 999, f(x) values decline from 0 to ∞, while f(x) values increase from -2 →  -1.001 towards ∞.

As the value of (x) declines from left to right and increases from right to left towards -1, we may assume that f(x) reaches - ∞ and + ∞, accordingly.

∴

$\mathbf{ \lim_{x \to \-1} \dfrac{x^2-7x}{x^2 -6x -7} = \infty}$