\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Evaluate the limit \(\displaystyle \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2}\)
The limit is an indeterminate form of the type \(\displaystyle \frac{\infty}{\infty}\). Using L'Hospitial's rule 3 times gives: \begin{equation*} \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} = \lim_{x \to -\infty}\frac{- 24 x^{2} - 18 x - 5}{24 x^{2} + 12 x + 2} = \lim_{x \to -\infty}\frac{- 6 \left(8 x + 3\right)}{12 \left(4 x + 1\right)} = \lim_{x \to -\infty}\frac{-48}{48} = -1 \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2}$
\soln{9cm}{The limit is an indeterminate form of the type $\frac{\infty}{\infty}$. Using L'Hospitial's rule 3 times gives: \begin{equation*} \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} = \lim_{x \to -\infty}\frac{- 24 x^{2} - 18 x - 5}{24 x^{2} + 12 x + 2} = \lim_{x \to -\infty}\frac{- 6 \left(8 x + 3\right)}{12 \left(4 x + 1\right)} = \lim_{x \to -\infty}\frac{-48}{48} = -1 \end{equation*}}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Evaluate the limit <img class="equation_image" title=" \displaystyle \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%208%20x%5E%7B3%7D%20-%209%20x%5E%7B2%7D%20-%205%20x%20-%201%7D%7B8%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%202%20x%20%2B%202%7D%20" alt="LaTeX: \displaystyle \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} " data-equation-content=" \displaystyle \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} " /> </p> </p><p> <p>The limit is an indeterminate form of the type <img class="equation_image" title=" \displaystyle \frac{\infty}{\infty} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%5Cinfty%7D%7B%5Cinfty%7D%20" alt="LaTeX: \displaystyle \frac{\infty}{\infty} " data-equation-content=" \displaystyle \frac{\infty}{\infty} " /> . Using L'Hospitial's rule 3 times gives: <img class="equation_image" title=" \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} = \lim_{x \to -\infty}\frac{- 24 x^{2} - 18 x - 5}{24 x^{2} + 12 x + 2} = \lim_{x \to -\infty}\frac{- 6 \left(8 x + 3\right)}{12 \left(4 x + 1\right)} = \lim_{x \to -\infty}\frac{-48}{48} = -1 " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%208%20x%5E%7B3%7D%20-%209%20x%5E%7B2%7D%20-%205%20x%20-%201%7D%7B8%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%202%20x%20%2B%202%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%2024%20x%5E%7B2%7D%20-%2018%20x%20-%205%7D%7B24%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%202%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%206%20%5Cleft%288%20x%20%2B%203%5Cright%29%7D%7B12%20%5Cleft%284%20x%20%2B%201%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-48%7D%7B48%7D%20%3D%20-1%20%20" alt="LaTeX: \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} = \lim_{x \to -\infty}\frac{- 24 x^{2} - 18 x - 5}{24 x^{2} + 12 x + 2} = \lim_{x \to -\infty}\frac{- 6 \left(8 x + 3\right)}{12 \left(4 x + 1\right)} = \lim_{x \to -\infty}\frac{-48}{48} = -1 " data-equation-content=" \lim_{x \to -\infty}\frac{- 8 x^{3} - 9 x^{2} - 5 x - 1}{8 x^{3} + 6 x^{2} + 2 x + 2} = \lim_{x \to -\infty}\frac{- 24 x^{2} - 18 x - 5}{24 x^{2} + 12 x + 2} = \lim_{x \to -\infty}\frac{- 6 \left(8 x + 3\right)}{12 \left(4 x + 1\right)} = \lim_{x \to -\infty}\frac{-48}{48} = -1 " /> </p> </p>