Q. 37 - Calculus

Question

Calculate each limit in Exercises 35-80.

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

Step-by-Step Solution

Answer

The limit of the function is $\underset{x\to \infty }{\lim }\left(2x^{-\frac{4}{3}}\right)=0$.

Step 1. Given information.

We have:

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

Step 2. Solve the limits.

The function is $\underset{x\to \infty }{\lim }\left(2x^{-\frac{4}{3}}\right)$.

$$\begin{gathered} \underset{x\to a}{\lim }\left[c·f\left(x\right)\right]=c·\underset{x\to a}{\lim }f\left(x\right) \\ =2·\underset{x\to \infty \hspace{0.17em}}{\lim }\left(x^{-\frac{4}{3}}\right) \\ \underset{x\to a}{\lim }{\left[f\left(x\right)\right]}^b={\left[\underset{x\to a}{\lim }f\left(x\right)\right]}^b \\ =2{\left(\underset{x\to \infty \hspace{0.17em}}{\lim }\left(x\right)\right)}^{-\frac{4}{3}} \\ =2{\infty }^{-\frac{4}{3}} \\ =0 \end{gathered}$$