Question

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

$\underset{x\to 1}{\lim }\frac{x-1}{x^3-1}$

Short answer

The value of limit of the given function is $\frac{1}{3}$.

Step-by-step solution

Step 1. Given information.

The limit of function here given is,

$\underset{x\to 1}{\lim }\frac{x-1}{x^3-1}$

Step 2. Formula used.

The algebraic formula will be used here to simplify.

That is,$a^3-b^3=(a-b)(a^2+ab+b^2)$

Step 3. Simplification.

Given, $\underset{x\to 1}{\lim }\frac{x-1}{x^3-1}$

Using the formula we get,

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

Step 4. Put the value.

Now, according to given table put the value of in the function $\frac{1}{(x^2+x+1)}$

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

If$x=0.9$, then$f\left(x\right)=0.36900$

If$x=0.99$, then$f\left(x\right)=0.33668$

If$x=0.999$, then$f\left(x\right)=0.33366$

If $x=1.001$, then$f\left(x\right)=0.33300$

If $x=1.01$, then $f\left(x\right)=0.30293$

If$x=1.1$, then$f\left(x\right)=0.30211$

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,

$\underset{x\to 1}{\lim }\frac{x-1}{x^3-1}=0.33=\frac{1}{3}$