Q. 60 - Calculus

Question

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

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

Step-by-Step Solution

Answer

$\underset{k\to \infty }{\lim }\left(\sqrt{k^3+k+1}-\sqrt{k^3-k-1}\right)=0$

Step 1. Given information

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

Step 2. Calculate integrate

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

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

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

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

$\underset{k\to \infty }{\lim }\left(\sqrt{k^3+k+1}-\sqrt{k^3-k-1}\right)=0$