Question

In Problems 7–16, use a table to find the indicated limit.

$12.\underset{x\to 3}{\lim }=\frac{x^2-9}{x^2-3x}$.

Short answer

$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-3x}=2$.

Step-by-step solution

Step 1. Given information.

Given function is $f\left(x\right)=\frac{x^2-9}{x^2-3x}$ whose limit tend to 3 need to be found.

Step 2. The table is as follows.

Compare the given limit to.
Choose values of x close to 3, arbitrarily starting with 2.99 then we select additional numbers that get closer to 3, but remain less than 3. Next choose values of x greater than 3, staring with 3.01, which get closer to3.

Finally evaluate f(x) at each choice to obtain the required Table.

As x gets closer to 3 then f(x) gets closer to 2.

So,$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-3x}=2$.