Question

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

$10.\underset{x\to 0}{\lim }\frac{2-x}{x^2+4}$.

Short answer

$\underset{x\to 0}{\lim }\frac{2-x}{x^2+4}=0.5$.

Step-by-step solution

Step 1. Given information.

A limit is given as$\underset{x\to c}{\lim }\frac{2-x}{x^2+4}$.

Step 2. Form the table.

Choose values of x close to 0, arbitrarily starting with -0.01then we select additional numbers that get closer to 0, but remain less than 0. Next choose values of x greater than 0, staring with 0.01, which get closer to 0.

Finally evaluate $f\left(x\right)=\frac{2-x}{x^2+4}$at each choice to obtain the below Table :

As x gets closer to 0 then f(x) gets closer to 0.5.

So,$\underset{x\to 0}{\lim }\frac{2-x}{x^2+4}=0.5$.