Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}.$

Short answer

The answer is$\frac{13}{28}.$

Step-by-step solution

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f\left(x\right)=\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$

Put $x=3$in the numerator we get,

$x^3-3x^2+4x-12=27-3\left(9\right)+4\left(3\right)-12=27-27+12-12=0.$

Put $x=3$in the denominator we get,

$x^4-3x^3+x-3=81-3\left(27\right)+3-3=81-81+3-3=0.$

Since both numerator and denominator gives 0 means they both have $x=3$as their common root.

Step. 2 Factorizing

Numerator , $x^3-3x^2+4x-12=x^2(x-3)+4(x-3)=(x^2+4)(x-3).$

Denominator,$x^4-3x^3+x-3=x^3(x-3)+1(x-3)=(x^3+1)(x-3).$

So,

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}=\underset{x\to 3}{\lim }\frac{(x^2+4)(x-3)}{(x^3+1)(x-3)} \\ =\underset{x\to 3}{\lim }\frac{x^2+4}{x^3+1}. \end{gathered}$$

Now we can put the limit directly.

Step. 3 Final calculation of the limit

$\underset{x\to 3}{\lim }\frac{x^2+4}{x^3+1}=\frac{9+4}{27+1}=\frac{13}{28}.$