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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6}\)
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{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} = \lim_{x \to -\infty}\frac{- 3 x^{2} + 16 x + 8}{9 x^{2} - 6 x - 8} = \lim_{x \to -\infty}\frac{2 \left(8 - 3 x\right)}{6 \left(3 x - 1\right)} = \lim_{x \to -\infty}\frac{-6}{18} = - \frac{1}{3} \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6}$
\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{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} = \lim_{x \to -\infty}\frac{- 3 x^{2} + 16 x + 8}{9 x^{2} - 6 x - 8} = \lim_{x \to -\infty}\frac{2 \left(8 - 3 x\right)}{6 \left(3 x - 1\right)} = \lim_{x \to -\infty}\frac{-6}{18} = - \frac{1}{3} \end{equation*}}
\end{question}
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\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{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%20x%5E%7B3%7D%20%2B%208%20x%5E%7B2%7D%20%2B%208%20x%20-%201%7D%7B3%20x%5E%7B3%7D%20-%203%20x%5E%7B2%7D%20-%208%20x%20-%206%7D%20" alt="LaTeX: \displaystyle \lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} " data-equation-content=" \displaystyle \lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} " /> </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{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} = \lim_{x \to -\infty}\frac{- 3 x^{2} + 16 x + 8}{9 x^{2} - 6 x - 8} = \lim_{x \to -\infty}\frac{2 \left(8 - 3 x\right)}{6 \left(3 x - 1\right)} = \lim_{x \to -\infty}\frac{-6}{18} = - \frac{1}{3} " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%20x%5E%7B3%7D%20%2B%208%20x%5E%7B2%7D%20%2B%208%20x%20-%201%7D%7B3%20x%5E%7B3%7D%20-%203%20x%5E%7B2%7D%20-%208%20x%20-%206%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%203%20x%5E%7B2%7D%20%2B%2016%20x%20%2B%208%7D%7B9%20x%5E%7B2%7D%20-%206%20x%20-%208%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B2%20%5Cleft%288%20-%203%20x%5Cright%29%7D%7B6%20%5Cleft%283%20x%20-%201%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-6%7D%7B18%7D%20%3D%20-%20%5Cfrac%7B1%7D%7B3%7D%20%20" alt="LaTeX: \lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} = \lim_{x \to -\infty}\frac{- 3 x^{2} + 16 x + 8}{9 x^{2} - 6 x - 8} = \lim_{x \to -\infty}\frac{2 \left(8 - 3 x\right)}{6 \left(3 x - 1\right)} = \lim_{x \to -\infty}\frac{-6}{18} = - \frac{1}{3} " data-equation-content=" \lim_{x \to -\infty}\frac{- x^{3} + 8 x^{2} + 8 x - 1}{3 x^{3} - 3 x^{2} - 8 x - 6} = \lim_{x \to -\infty}\frac{- 3 x^{2} + 16 x + 8}{9 x^{2} - 6 x - 8} = \lim_{x \to -\infty}\frac{2 \left(8 - 3 x\right)}{6 \left(3 x - 1\right)} = \lim_{x \to -\infty}\frac{-6}{18} = - \frac{1}{3} " /> </p> </p>