Question

find the limit.

$\underset{t\to \infty }{\lim }\frac{8t^3+t}{\left(2t-1\right)\left(2t^2+1\right)}$

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,

$\underset{t\to \infty }{\lim }\frac{8t^3+t}{(2t-1)(2t^2+1)}$

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,

$\begin{array}{c}\underset{t\to \infty }{\lim }\frac{8t^3+t}{(2t-1)(2t^2+1)}\\ =\underset{t\to \infty }{\lim }\frac{8t^3+t}{4t^3+2t-2t^2-1}\end{array}$

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

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

$\begin{array}{c}\underset{t\to \infty }{\lim }\frac{8t^3+t}{4t^3+2t-2t^2-1}=\frac{\underset{t\to \infty }{\lim }8+\underset{t\to \infty }{\lim }\frac{1}{t^2}}{\underset{t\to \infty }{\lim }4+\underset{t\to \infty }{\lim }\frac{2}{t^2}-\underset{t\to \infty }{\lim }\frac{2}{t}-\underset{t\to \infty }{\lim }\frac{1}{t^3}}\\ \end{array}$

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

$\begin{array}{c}=\frac{8+0}{4+0-0-0}\\ =\frac{8}{4}\\ =2\end{array}$

Therefore, our final answer is $2.$