Question

find the limit $\underset{x\to 2}{\lim }\frac{x^3-8}{x^3-2x^2+4x-8}$

Short answer

$\underset{x\to 2}{\lim }\frac{x^3-8}{x^3-2x^2+4x-8}=\frac{3}{2}$

Step-by-step solution

Step 1. Given information 

$\underset{x\to 2}{\lim }\frac{x^3-8}{x^3-2x^2+4x-8}$

Step 2. limit 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) and in the domain of f(x)= $\frac{x^3-8}{x^3-2x^2+4x-8}$2 is not included. So we have to simplify the f(x) for finding the limit at 2.

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^3-8}{x^3-2x^2+4x-8}=\underset{x\to 2}{\lim }\frac{(x-2)\left(x^2+2x+4\right)}{\left(x^2+4\right)(x-2)} \\ =\underset{x\to 2}{\lim }\frac{x^2+2x+4}{x^2+4} \\ =\frac{(2{)}^2+2(2)+4}{(2{)}^2+4}=\frac{3}{2} \end{gathered}$$