Q. 57 - Calculus

Question

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

$\underset{k\to \infty }{\lim }\frac{k-7}{3k+5}$

Step-by-Step Solution

Answer

$\underset{k\to \infty }{\lim }\frac{k-7}{3k+5}=\frac{1}{3}$

Step 1. Given information

$\underset{k\to \infty }{\lim }\frac{k-7}{3k+5}$

Step 2. Calculate the limit

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

$\underset{k\to \infty }{\lim }\frac{k-7}{3k+5}=\underset{x\to \infty \hspace{0.17em}}{\lim }\left(\frac{1-\frac{7}{x}}{3+\frac{5}{x}}\right)$

$\underset{k\to \infty }{\lim }\frac{k-7}{3k+5}=\frac{1}{3}$