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Questions: Algebra BusinessCalculus
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Evaluate the limit \(\displaystyle \lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12}\)
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 -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} = \lim_{x \to -3}\frac{10 x + 5}{16 x + 28} = \frac{10 (-3) + 5}{16 (-3) + 28} = \frac{5}{4} \end{equation*}
\begin{question}Evaluate the limit $\lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12}$
\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 -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} = \lim_{x \to -3}\frac{10 x + 5}{16 x + 28} = \frac{10 (-3) + 5}{16 (-3) + 28} = \frac{5}{4} \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 -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-3%7D%5Cfrac%7B5%20x%5E%7B2%7D%20%2B%205%20x%20-%2030%7D%7B8%20x%5E%7B2%7D%20%2B%2028%20x%20%2B%2012%7D%20" alt="LaTeX: \displaystyle \lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} " data-equation-content=" \displaystyle \lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} " /> </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 -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} = \lim_{x \to -3}\frac{10 x + 5}{16 x + 28} = \frac{10 (-3) + 5}{16 (-3) + 28} = \frac{5}{4} " src="/equation_images/%20%20%5Clim_%7Bx%20%5Cto%20-3%7D%5Cfrac%7B5%20x%5E%7B2%7D%20%2B%205%20x%20-%2030%7D%7B8%20x%5E%7B2%7D%20%2B%2028%20x%20%2B%2012%7D%20%3D%20%5Clim_%7Bx%20%5Cto%20-3%7D%5Cfrac%7B10%20x%20%2B%205%7D%7B16%20x%20%2B%2028%7D%20%3D%20%5Cfrac%7B10%20%28-3%29%20%2B%205%7D%7B16%20%28-3%29%20%2B%2028%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D%20%20" alt="LaTeX: \lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} = \lim_{x \to -3}\frac{10 x + 5}{16 x + 28} = \frac{10 (-3) + 5}{16 (-3) + 28} = \frac{5}{4} " data-equation-content=" \lim_{x \to -3}\frac{5 x^{2} + 5 x - 30}{8 x^{2} + 28 x + 12} = \lim_{x \to -3}\frac{10 x + 5}{16 x + 28} = \frac{10 (-3) + 5}{16 (-3) + 28} = \frac{5}{4} " /> </p> </p>