Question

let $f(x) = \frac{x^2 + x - 6}{x - 2}$. fill out the tables to evaluate $\lim\limits_{x\to 2} f(x)$.

$x$	$f(x)$		$x$	$f(x)$
1.99			2.01
1.999			2.001

\bigcirc 5
\bigcirc 0
\bigcirc 2
\bigcirc 3
\bigcirc the limit does not exist.

Explanation:

Step1: Simplify the function

First, factor the numerator \(x^2 + x - 6\). We know that \(x^2 + x - 6=(x + 3)(x - 2)\). So the function \(f(x)=\frac{x^2+x - 6}{x - 2}=\frac{(x + 3)(x - 2)}{x - 2}\) (for \(x
eq2\)). Then we can cancel out the \(x - 2\) terms (since \(x
eq2\) when we are taking the limit as \(x
ightarrow2\)), so \(f(x)=x + 3\) for \(x
eq2\).

Step2: Evaluate the limit

To find \(\lim_{x
ightarrow2}f(x)\), we can use the simplified function \(f(x)=x + 3\) (since the limit as \(x
ightarrow2\) does not depend on the value of the function at \(x = 2\), only the values around \(x = 2\)). Substitute \(x = 2\) into \(x+3\), we get \(2 + 3=5\).

We can also check by plugging in the values from the table:

• For \(x = 1.9\): \(f(1.9)=1.9 + 3=4.9\)
• For \(x = 1.99\): \(f(1.99)=1.99+3 = 4.99\)
• For \(x = 1.999\): \(f(1.999)=1.999 + 3=4.999\)
• For \(x = 2.1\): \(f(2.1)=2.1+3 = 5.1\)
• For \(x = 2.01\): \(f(2.01)=2.01 + 3=5.01\)
• For \(x = 2.001\): \(f(2.001)=2.001+3 = 5.001\)

As \(x\) approaches \(2\) from both the left and the right, \(f(x)\) approaches \(5\).

Answer:

5