Question

find the limit $\underset{x\to 3}{\lim }\frac{x^4-3x^3+x-3}{x^3-3x^2+2x-6}$

Short answer

$\underset{x\to 3}{\lim }\frac{x^4-3x^3+x-3}{x^3-3x^2+2x-6}=\frac{28}{11}$

Step-by-step solution

Step 1. Given information 

$\underset{x\to 3}{\lim }\frac{x^4-3x^3+x-3}{x^3-3x^2+2x-6}$

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^4-3x^3+x-3}{x^3-3x^2+2x-6}$3 is not included So we have to simplify the f(x) for finding the limit at 3.

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{x^4-3x^3+x-3}{x^3-3x^2+2x-6}=\underset{x\to 3}{\lim }\frac{\left(x^3+1\right)(x-3)}{\left(x^2+2\right)(x-3)} \\ =\underset{x\to 3}{\lim }\frac{x^3+1}{x^2+2} \\ =\frac{(3{)}^3+1}{(3{)}^2+2}=\frac{28}{11} \end{gathered}$$