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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9}\)
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{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} = \lim_{x \to \infty}\frac{27 x^{2} - 10 x + 9}{- 27 x^{2} - 12 x + 2} = \lim_{x \to \infty}\frac{2 \left(27 x - 5\right)}{- 6 \left(9 x + 2\right)} = \lim_{x \to \infty}\frac{54}{-54} = -1 \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9}$
\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{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} = \lim_{x \to \infty}\frac{27 x^{2} - 10 x + 9}{- 27 x^{2} - 12 x + 2} = \lim_{x \to \infty}\frac{2 \left(27 x - 5\right)}{- 6 \left(9 x + 2\right)} = \lim_{x \to \infty}\frac{54}{-54} = -1 \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{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B9%20x%5E%7B3%7D%20-%205%20x%5E%7B2%7D%20%2B%209%20x%20%2B%205%7D%7B-%209%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%202%20x%20%2B%209%7D%20" alt="LaTeX: \displaystyle \lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} " data-equation-content=" \displaystyle \lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 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{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} = \lim_{x \to \infty}\frac{27 x^{2} - 10 x + 9}{- 27 x^{2} - 12 x + 2} = \lim_{x \to \infty}\frac{2 \left(27 x - 5\right)}{- 6 \left(9 x + 2\right)} = \lim_{x \to \infty}\frac{54}{-54} = -1 " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B9%20x%5E%7B3%7D%20-%205%20x%5E%7B2%7D%20%2B%209%20x%20%2B%205%7D%7B-%209%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%202%20x%20%2B%209%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B27%20x%5E%7B2%7D%20-%2010%20x%20%2B%209%7D%7B-%2027%20x%5E%7B2%7D%20-%2012%20x%20%2B%202%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%20%5Cleft%2827%20x%20-%205%5Cright%29%7D%7B-%206%20%5Cleft%289%20x%20%2B%202%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B54%7D%7B-54%7D%20%3D%20-1%20%20" alt="LaTeX: \lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} = \lim_{x \to \infty}\frac{27 x^{2} - 10 x + 9}{- 27 x^{2} - 12 x + 2} = \lim_{x \to \infty}\frac{2 \left(27 x - 5\right)}{- 6 \left(9 x + 2\right)} = \lim_{x \to \infty}\frac{54}{-54} = -1 " data-equation-content=" \lim_{x \to \infty}\frac{9 x^{3} - 5 x^{2} + 9 x + 5}{- 9 x^{3} - 6 x^{2} + 2 x + 9} = \lim_{x \to \infty}\frac{27 x^{2} - 10 x + 9}{- 27 x^{2} - 12 x + 2} = \lim_{x \to \infty}\frac{2 \left(27 x - 5\right)}{- 6 \left(9 x + 2\right)} = \lim_{x \to \infty}\frac{54}{-54} = -1 " /> </p> </p>