Question

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. Domain of the function

We can find limit of rational function by $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$ if c is in the domain of f(x) .So to find the domain of f(x)= $\frac{x-1}{x^3-1}$we have to put denominator =0

$$\begin{gathered} x^3-1=0 \\ x=1 \end{gathered}$$

So the domain of f(x) is all real number except 1.

Step 3. limit of the function 

As c=1 in our limit so we have to simplify the f(x).

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x-1}{x^3-1}=\underset{x\to 1}{\lim }\frac{x-1}{(x-1)\left(x^2+x+1\right)} \\ =\underset{x\to 1}{\lim }\frac{1}{x^2+x+1} \\ =\frac{1}{(1{)}^2+(1)+1}=\frac{1}{3} \end{gathered}$$