Question

find the limit.

$\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{{\left(r+1\right)}^3}$

Short answer

The value of limit of the given function is $2.$

Step-by-step solution

Step 1. Given information.

The limit of function is given here,

$\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{{(r+1)}^3}$

Step 2. Formula used.

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

We apply the formula:

$\begin{array}{l}\underset{x\to \infty }{\lim }\frac{1}{x}=0\\ \underset{x\to \infty }{\lim }\frac{1}{x^k}=0\\ \text{Invalid <m:msup> element}=a^3+b^3+3ab(a+b)\end{array}$

Step 3. Simplification.

Given,

$\begin{array}{c}\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{{(r+1)}^3}\\ =\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{r^3+1+3r(r+1)}\\ =\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{r^3+1+3r^2+3r}\end{array}$

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

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

$\begin{array}{c}\underset{r\to \infty }{\lim }\frac{4r^3-r^2}{r^3+1+3r^2+3r}=\underset{r\to \infty }{\lim }\left(\frac{4-\frac{1}{r}}{1+1+\frac{3}{r}+\frac{3}{r^2}}\right)\\ =\left(\frac{\underset{r\to \infty }{\lim }4-\underset{r\to \infty }{\lim }\frac{1}{r}}{\underset{r\to \infty }{\lim }1+\underset{r\to \infty }{\lim }1+\underset{r\to \infty }{\lim }\frac{3}{r}+\underset{r\to \infty }{\lim }\frac{3}{r^2}}\right)\\ \end{array}$

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

$\begin{array}{c}=\frac{4-0}{1+1+0+0}\\ =\frac{4}{2}\\ =2\end{array}$

Therefore, our final answer is $2.$