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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5}\)
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} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} = \lim_{x \to \infty}\frac{6 x^{2} - 2 x - 7}{6 x^{2} - 8 x + 1} = \lim_{x \to \infty}\frac{2 \left(6 x - 1\right)}{4 \left(3 x - 2\right)} = \lim_{x \to \infty}\frac{12}{12} = 1 \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5}$
\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} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} = \lim_{x \to \infty}\frac{6 x^{2} - 2 x - 7}{6 x^{2} - 8 x + 1} = \lim_{x \to \infty}\frac{2 \left(6 x - 1\right)}{4 \left(3 x - 2\right)} = \lim_{x \to \infty}\frac{12}{12} = 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{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%207%20x%20%2B%207%7D%7B2%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20%2B%20x%20%2B%205%7D%20" alt="LaTeX: \displaystyle \lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} " data-equation-content=" \displaystyle \lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} " /> </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} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} = \lim_{x \to \infty}\frac{6 x^{2} - 2 x - 7}{6 x^{2} - 8 x + 1} = \lim_{x \to \infty}\frac{2 \left(6 x - 1\right)}{4 \left(3 x - 2\right)} = \lim_{x \to \infty}\frac{12}{12} = 1 " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%207%20x%20%2B%207%7D%7B2%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20%2B%20x%20%2B%205%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B6%20x%5E%7B2%7D%20-%202%20x%20-%207%7D%7B6%20x%5E%7B2%7D%20-%208%20x%20%2B%201%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%20%5Cleft%286%20x%20-%201%5Cright%29%7D%7B4%20%5Cleft%283%20x%20-%202%5Cright%29%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B12%7D%7B12%7D%20%3D%201%20%20" alt="LaTeX: \lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} = \lim_{x \to \infty}\frac{6 x^{2} - 2 x - 7}{6 x^{2} - 8 x + 1} = \lim_{x \to \infty}\frac{2 \left(6 x - 1\right)}{4 \left(3 x - 2\right)} = \lim_{x \to \infty}\frac{12}{12} = 1 " data-equation-content=" \lim_{x \to \infty}\frac{2 x^{3} - x^{2} - 7 x + 7}{2 x^{3} - 4 x^{2} + x + 5} = \lim_{x \to \infty}\frac{6 x^{2} - 2 x - 7}{6 x^{2} - 8 x + 1} = \lim_{x \to \infty}\frac{2 \left(6 x - 1\right)}{4 \left(3 x - 2\right)} = \lim_{x \to \infty}\frac{12}{12} = 1 " /> </p> </p>