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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8}\)
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{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} = \lim_{x \to -\infty}\frac{- 6 x^{2} + 10 x + 5}{- 15 x^{2} - 10 x + 5} = \lim_{x \to -\infty}\frac{2 \left(5 - 6 x\right)}{- 10 \left(3 x + 1\right)} = \lim_{x \to -\infty}\frac{-12}{-30} = \frac{2}{5} \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8}$
\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{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} = \lim_{x \to -\infty}\frac{- 6 x^{2} + 10 x + 5}{- 15 x^{2} - 10 x + 5} = \lim_{x \to -\infty}\frac{2 \left(5 - 6 x\right)}{- 10 \left(3 x + 1\right)} = \lim_{x \to -\infty}\frac{-12}{-30} = \frac{2}{5} \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{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%202%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%205%20x%20-%208%7D%7B-%205%20x%5E%7B3%7D%20-%205%20x%5E%7B2%7D%20%2B%205%20x%20-%208%7D%20" alt="LaTeX: \displaystyle \lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} " data-equation-content=" \displaystyle \lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} " /> </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{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} = \lim_{x \to -\infty}\frac{- 6 x^{2} + 10 x + 5}{- 15 x^{2} - 10 x + 5} = \lim_{x \to -\infty}\frac{2 \left(5 - 6 x\right)}{- 10 \left(3 x + 1\right)} = \lim_{x \to -\infty}\frac{-12}{-30} = \frac{2}{5} " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%202%20x%5E%7B3%7D%20%2B%205%20x%5E%7B2%7D%20%2B%205%20x%20-%208%7D%7B-%205%20x%5E%7B3%7D%20-%205%20x%5E%7B2%7D%20%2B%205%20x%20-%208%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-%206%20x%5E%7B2%7D%20%2B%2010%20x%20%2B%205%7D%7B-%2015%20x%5E%7B2%7D%20-%2010%20x%20%2B%205%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B2%20%5Cleft%285%20-%206%20x%5Cright%29%7D%7B-%2010%20%5Cleft%283%20x%20%2B%201%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-%5Cinfty%7D%5Cfrac%7B-12%7D%7B-30%7D%20%3D%20%5Cfrac%7B2%7D%7B5%7D%20%20" alt="LaTeX: \lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} = \lim_{x \to -\infty}\frac{- 6 x^{2} + 10 x + 5}{- 15 x^{2} - 10 x + 5} = \lim_{x \to -\infty}\frac{2 \left(5 - 6 x\right)}{- 10 \left(3 x + 1\right)} = \lim_{x \to -\infty}\frac{-12}{-30} = \frac{2}{5} " data-equation-content=" \lim_{x \to -\infty}\frac{- 2 x^{3} + 5 x^{2} + 5 x - 8}{- 5 x^{3} - 5 x^{2} + 5 x - 8} = \lim_{x \to -\infty}\frac{- 6 x^{2} + 10 x + 5}{- 15 x^{2} - 10 x + 5} = \lim_{x \to -\infty}\frac{2 \left(5 - 6 x\right)}{- 10 \left(3 x + 1\right)} = \lim_{x \to -\infty}\frac{-12}{-30} = \frac{2}{5} " /> </p> </p>