Question

find the limit.

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

Short answer

The value of limit of the given function is $0.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

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

Step 2. Formula used.

Here, in the given question we apply limit of a quotient and limit of sum, difference.

After it we apply the formula:

$\underset{x\to -\infty }{\lim }\frac{1}{x}=0$

$\underset{x\to -\infty }{\lim }\frac{1}{x^k}=0$

Step 3. Simplification.

Given,

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

Here, first we divide the numerator and denominator by the greatest power of $x$given in the question then we apply limit of constant multiple.

So, divide the numerator and denominator by $x^3$.

$\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{x^2+2}{x^3+x+1}=\underset{x\to -\infty }{\lim }\left(\frac{\frac{1}{x}+\frac{2}{x^3}}{1+\frac{1}{x^2}+\frac{1}{x^3}}\right)\\ =\left(\frac{\underset{x\to -\infty }{\lim }\frac{1}{x}+\underset{x\to -\infty }{\lim }\frac{2}{x^3}}{\underset{x\to -\infty }{\lim }1+\underset{x\to -\infty }{\lim }\frac{1}{x^2}+\underset{x\to -\infty }{\lim }\frac{1}{x^3}}\right)\end{array}$

Now, we apply the value of $x$which is equals to $\infty .$

$\begin{array}{c}=\frac{0+0}{1+0+0}\\ =0\end{array}$

Therefore, our final answer is $0.$