Q. 58 - Calculus

Question

Evaluate the limits in Exercises 55–60. Use the theorems in this section to justify each step of your work.  

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

Step-by-Step Solution

Answer

$\underset{k\to \infty }{\lim }\left(\frac{k^3+1}{k^2}\right)=\infty$

Step 1. Given information

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

Step 2. Calculate the limit

$\underset{k\to \infty }{\lim }\left(\frac{k^3+1}{k^2}\right)=\underset{x\to \infty \hspace{0.17em}}{\lim }\left(x+\frac{1}{x^2}\right)$

$\underset{x\to a}{\lim }\left[f\left(x\right)\pm g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\pm \underset{x\to a}{\lim }g\left(x\right)$

$\underset{k\to \infty }{\lim }\left(\frac{k^3+1}{k^2}\right)=\infty$