Question

Complete the table of values (to five decimal places), and use the table to estimate the value of the limit.

$\underset{x\to 2}{\lim }\frac{x-2}{x^2+x-6}$

Short answer

The value of limit of the given function is$\frac{1}{5}$.

Step-by-step solution

Step 1. Given information.

The limit of function here given is,

$\underset{x\to 2}{\lim }\frac{x-2}{x^2+x-6}$

Step 2. Simplification.

Given, $\underset{x\to 2}{\lim }\frac{x-2}{x^2+x-6}$

Using the factorisation method we get,

$\begin{array}{c}\underset{x\to 2}{\lim }\frac{x-2}{x^2+3x-2x-6}\\ \underset{x\to 2}{\lim }\frac{x-2}{x(x+3)-2(x+3)}\\ \underset{x\to 2}{\lim }\frac{(x-2)}{(x+3)(x-2)}\\ \underset{x\to 2}{\lim }\frac{1}{(x+3)}\end{array}$

Step 3. Put the value.

Now, according to given table put the value of $x$in the function$\frac{1}{(x+3)}$

And get the value of$f\left(x\right)$

If $x=1.9$,then $f\left(x\right)=0.20408$

If$x=1.99$,then$f\left(x\right)=0.20040$

If$x=1.999$, then$f\left(x\right)=0.20004$

If $x=2.001$,then$f\left(x\right)=0.19996$

If $x=2.01$,then$f\left(x\right)=0.19960$

If$x=2.1$,then$f\left(x\right)=0.19607$

Step 5. Complete the table.

Now, the completed table is below,

Therefore, on the basis of the values in the table we make the guess that,

$\underset{x\to 2}{\lim }\frac{x-2}{x^2+x-6}=0.20=\frac{1}{5}$