Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to -1}{\lim }\frac{x^3+x^2+3x+3}{x^4+x^3+2x+2}$.

Short answer

The answer is 4.

Step-by-step solution

Step. 1 Given Information

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

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

Put $x=-1$in the numerator we get,

$x^3+x^2+3x+3={(-1)}^3+{(-1)}^2+3(-1)+3=-1+1-3+3=0.$

Put $x=-1$in the denominator we get,

$x^4+x^3+2x+2={(-1)}^4+{(-1)}^3+2(-1)+2=1-1-2+2=0.$

Since both numerator and denominator gives 0 means they both have $x=-1$as their common root.

Step. 2 Factorizing

Denominator,$x^4+x^3+2x+2=x^3(x+1)+2(x+1)=(x^3+2)(x+1)$.

Numerator ,$x^3+x^2+3x+3=x^2(x+1)+3(x+1)=(x^2+3)(x+1).$

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit 

$\underset{x\to -1}{\lim }\frac{x^2+3}{x^3+2}=\frac{1+3}{-1+2}=\frac{4}{1}=4.$