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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25}\)
The limit is an indeterminate form of the type \(\displaystyle \frac{0}{0}\). Using L'Hospitial's rule and then the evaluation theorem gives: \begin{equation*} \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} = \lim_{x \to 5}\frac{6 x - 36}{8 x - 15} = \frac{6 (5) - 36}{8 (5) - 15} = - \frac{6}{25} \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25}$
\soln{9cm}{The limit is an indeterminate form of the type $\frac{0}{0}$. Using L'Hospitial's rule and then the evaluation theorem gives: \begin{equation*} \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} = \lim_{x \to 5}\frac{6 x - 36}{8 x - 15} = \frac{6 (5) - 36}{8 (5) - 15} = - \frac{6}{25} \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 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%205%7D%5Cfrac%7B3%20x%5E%7B2%7D%20-%2036%20x%20%2B%20105%7D%7B4%20x%5E%7B2%7D%20-%2015%20x%20-%2025%7D%20" alt="LaTeX: \displaystyle \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} " data-equation-content=" \displaystyle \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} " /> </p> </p><p> <p>The limit is an indeterminate form of the type <img class="equation_image" title=" \displaystyle \frac{0}{0} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B0%7D%7B0%7D%20" alt="LaTeX: \displaystyle \frac{0}{0} " data-equation-content=" \displaystyle \frac{0}{0} " /> . Using L'Hospitial's rule and then the evaluation theorem gives: <img class="equation_image" title=" \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} = \lim_{x \to 5}\frac{6 x - 36}{8 x - 15} = \frac{6 (5) - 36}{8 (5) - 15} = - \frac{6}{25} " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%205%7D%5Cfrac%7B3%20x%5E%7B2%7D%20-%2036%20x%20%2B%20105%7D%7B4%20x%5E%7B2%7D%20-%2015%20x%20-%2025%7D%20%3D%20%5Clim_%7Bx%20%5Cto%205%7D%5Cfrac%7B6%20x%20-%2036%7D%7B8%20x%20-%2015%7D%20%3D%20%5Cfrac%7B6%20%285%29%20-%2036%7D%7B8%20%285%29%20-%2015%7D%20%3D%20-%20%5Cfrac%7B6%7D%7B25%7D%20%20" alt="LaTeX: \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} = \lim_{x \to 5}\frac{6 x - 36}{8 x - 15} = \frac{6 (5) - 36}{8 (5) - 15} = - \frac{6}{25} " data-equation-content=" \lim_{x \to 5}\frac{3 x^{2} - 36 x + 105}{4 x^{2} - 15 x - 25} = \lim_{x \to 5}\frac{6 x - 36}{8 x - 15} = \frac{6 (5) - 36}{8 (5) - 15} = - \frac{6}{25} " /> </p> </p>