Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is 2.

Step-by-step solution

Step. 1 Given Information

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

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

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

$x^3+2x^2+x=-1+2-1=0.$

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

$x^4+x^3+2x+2=1-1-2+2=0.$

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

Step. 2 Factorizing

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

Denominator,role="math" localid="1647142064867" $x^4+x^3+2x+2=x^3(x+1)+2(x+1)=(x^3+2)(x+1).$

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit

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