Question

Evaluate the limit if it exists.

$\underset{x\to 1}{\lim }\frac{x^3-1}{x^2-1}$

Short answer

The value of limit of the given function is$\frac{3}{2}.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{x\to 1}{\lim }\frac{x^3-1}{x^2-1}$

Step 2. Formula used.

The given function is solved by applying the formula and then put the value.

Formula:$\begin{array}{c}\left(a^3-b^3\right)=(a-b)(a^2+ab+b^2)\\ (a^2-b^2)=(a+b)(a-b)\end{array}$

Step 3. Simplification.

Given, $\underset{x\to 1}{\lim }\frac{x^3-1}{x^2-1}$

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

Therefore, our final answer is$\frac{3}{2}.$