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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4}\)
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{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} = \lim_{x \to \infty}\frac{- 12 x^{2} + 2 x + 4}{6 x^{2} + 18 x + 7} = \lim_{x \to \infty}\frac{2 \left(1 - 12 x\right)}{6 \left(2 x + 3\right)} = \lim_{x \to \infty}\frac{-24}{12} = -2 \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4}$
\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{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} = \lim_{x \to \infty}\frac{- 12 x^{2} + 2 x + 4}{6 x^{2} + 18 x + 7} = \lim_{x \to \infty}\frac{2 \left(1 - 12 x\right)}{6 \left(2 x + 3\right)} = \lim_{x \to \infty}\frac{-24}{12} = -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{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%204%20x%5E%7B3%7D%20%2B%20x%5E%7B2%7D%20%2B%204%20x%20-%201%7D%7B2%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%207%20x%20%2B%204%7D%20" alt="LaTeX: \displaystyle \lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} " data-equation-content=" \displaystyle \lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} " /> </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{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} = \lim_{x \to \infty}\frac{- 12 x^{2} + 2 x + 4}{6 x^{2} + 18 x + 7} = \lim_{x \to \infty}\frac{2 \left(1 - 12 x\right)}{6 \left(2 x + 3\right)} = \lim_{x \to \infty}\frac{-24}{12} = -2 " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%204%20x%5E%7B3%7D%20%2B%20x%5E%7B2%7D%20%2B%204%20x%20-%201%7D%7B2%20x%5E%7B3%7D%20%2B%209%20x%5E%7B2%7D%20%2B%207%20x%20%2B%204%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-%2012%20x%5E%7B2%7D%20%2B%202%20x%20%2B%204%7D%7B6%20x%5E%7B2%7D%20%2B%2018%20x%20%2B%207%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%20%5Cleft%281%20-%2012%20x%5Cright%29%7D%7B6%20%5Cleft%282%20x%20%2B%203%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B-24%7D%7B12%7D%20%3D%20-2%20%20" alt="LaTeX: \lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} = \lim_{x \to \infty}\frac{- 12 x^{2} + 2 x + 4}{6 x^{2} + 18 x + 7} = \lim_{x \to \infty}\frac{2 \left(1 - 12 x\right)}{6 \left(2 x + 3\right)} = \lim_{x \to \infty}\frac{-24}{12} = -2 " data-equation-content=" \lim_{x \to \infty}\frac{- 4 x^{3} + x^{2} + 4 x - 1}{2 x^{3} + 9 x^{2} + 7 x + 4} = \lim_{x \to \infty}\frac{- 12 x^{2} + 2 x + 4}{6 x^{2} + 18 x + 7} = \lim_{x \to \infty}\frac{2 \left(1 - 12 x\right)}{6 \left(2 x + 3\right)} = \lim_{x \to \infty}\frac{-24}{12} = -2 " /> </p> </p>