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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9}\)
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{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} = \lim_{x \to \infty}\frac{- 18 x^{2} + 16 x + 5}{9 x^{2} + 14 x - 3} = \lim_{x \to \infty}\frac{4 \left(4 - 9 x\right)}{2 \left(9 x + 7\right)} = \lim_{x \to \infty}\frac{-36}{18} = -2 \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9}$
\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{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} = \lim_{x \to \infty}\frac{- 18 x^{2} + 16 x + 5}{9 x^{2} + 14 x - 3} = \lim_{x \to \infty}\frac{4 \left(4 - 9 x\right)}{2 \left(9 x + 7\right)} = \lim_{x \to \infty}\frac{-36}{18} = -2 \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{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%206%20x%5E%7B3%7D%20%2B%208%20x%5E%7B2%7D%20%2B%205%20x%20-%205%7D%7B3%20x%5E%7B3%7D%20%2B%207%20x%5E%7B2%7D%20-%203%20x%20-%209%7D%20" alt="LaTeX: \displaystyle \lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} " data-equation-content=" \displaystyle \lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} " /> </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{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} = \lim_{x \to \infty}\frac{- 18 x^{2} + 16 x + 5}{9 x^{2} + 14 x - 3} = \lim_{x \to \infty}\frac{4 \left(4 - 9 x\right)}{2 \left(9 x + 7\right)} = \lim_{x \to \infty}\frac{-36}{18} = -2 " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%206%20x%5E%7B3%7D%20%2B%208%20x%5E%7B2%7D%20%2B%205%20x%20-%205%7D%7B3%20x%5E%7B3%7D%20%2B%207%20x%5E%7B2%7D%20-%203%20x%20-%209%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%2018%20x%5E%7B2%7D%20%2B%2016%20x%20%2B%205%7D%7B9%20x%5E%7B2%7D%20%2B%2014%20x%20-%203%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B4%20%5Cleft%284%20-%209%20x%5Cright%29%7D%7B2%20%5Cleft%289%20x%20%2B%207%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-36%7D%7B18%7D%20%3D%20-2%20%20" alt="LaTeX: \lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} = \lim_{x \to \infty}\frac{- 18 x^{2} + 16 x + 5}{9 x^{2} + 14 x - 3} = \lim_{x \to \infty}\frac{4 \left(4 - 9 x\right)}{2 \left(9 x + 7\right)} = \lim_{x \to \infty}\frac{-36}{18} = -2 " data-equation-content=" \lim_{x \to \infty}\frac{- 6 x^{3} + 8 x^{2} + 5 x - 5}{3 x^{3} + 7 x^{2} - 3 x - 9} = \lim_{x \to \infty}\frac{- 18 x^{2} + 16 x + 5}{9 x^{2} + 14 x - 3} = \lim_{x \to \infty}\frac{4 \left(4 - 9 x\right)}{2 \left(9 x + 7\right)} = \lim_{x \to \infty}\frac{-36}{18} = -2 " /> </p> </p>