Q. 59 - Calculus

Question

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

$\underset{x\to \infty }{\lim }\sqrt{k^2+k}-k$

Step-by-Step Solution

Answer

$\underset{x\to \infty }{\lim }\sqrt{k^2+k}-k=\frac{1}{2}$

Step 1. Given information

$\underset{x\to \infty }{\lim }\sqrt{k^2+k}-k$

Step 2. Calculate the limit

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

$\underset{x\to a}{\lim }\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{{\lim }_{x\to a}f\left(x\right)}{{\lim }_{x\to a}g\left(x\right)},\hspace{0.17em} \underset{x\to a}{\lim }g\left(x\right)\ne 0$

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

$\underset{x\to \infty }{\lim }\sqrt{k^2+k}-k=\frac{1}{2}$