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Questions: Algebra BusinessCalculus
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Find the limit, if it exists \(\displaystyle \lim_{x \to -2 }\frac{ \left(x + 6\right)^{2} - 16 }{ x + 2 }\)
Expanding the numerator and simplifying gives: \begin{equation*} \frac{ x^{2} + 12 x + 36 - 16 }{ x + 2 } = \frac{ x^{2} + 12 x + 20 }{ x + 2 } = \frac{ \left(x + 2\right) \left(x + 10\right) }{ x + 2 } = x + 10 \end{equation*}The reduced function is continuous at \(\displaystyle x = -2\) and by the evaluation theorem \(\displaystyle \lim_{x \to -2 }x + 10 = 8\)
\begin{question}Find the limit, if it exists $\lim_{x \to -2 }\frac{ \left(x + 6\right)^{2} - 16 }{ x + 2 }$
\soln{9cm}{Expanding the numerator and simplifying gives: \begin{equation*} \frac{ x^{2} + 12 x + 36 - 16 }{ x + 2 } = \frac{ x^{2} + 12 x + 20 }{ x + 2 } = \frac{ \left(x + 2\right) \left(x + 10\right) }{ x + 2 } = x + 10 \end{equation*}The reduced function is continuous at $x = -2$ and by the evaluation theorem $\lim_{x \to -2 }x + 10 = 8$}
\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>Find the limit, if it exists <img class="equation_image" title=" \displaystyle \lim_{x \to -2 }\frac{ \left(x + 6\right)^{2} - 16 }{ x + 2 } " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-2%20%7D%5Cfrac%7B%20%5Cleft%28x%20%2B%206%5Cright%29%5E%7B2%7D%20-%2016%20%7D%7B%20x%20%2B%202%20%7D%20" alt="LaTeX: \displaystyle \lim_{x \to -2 }\frac{ \left(x + 6\right)^{2} - 16 }{ x + 2 } " data-equation-content=" \displaystyle \lim_{x \to -2 }\frac{ \left(x + 6\right)^{2} - 16 }{ x + 2 } " /> </p> </p><p> <p>Expanding the numerator and simplifying gives: <img class="equation_image" title=" \frac{ x^{2} + 12 x + 36 - 16 }{ x + 2 } = \frac{ x^{2} + 12 x + 20 }{ x + 2 } = \frac{ \left(x + 2\right) \left(x + 10\right) }{ x + 2 } = x + 10 " src="/equation_images/%20%20%5Cfrac%7B%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%2036%20-%2016%20%7D%7B%20x%20%2B%202%20%7D%20%3D%20%5Cfrac%7B%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%2020%20%7D%7B%20x%20%2B%202%20%7D%20%3D%20%20%5Cfrac%7B%20%5Cleft%28x%20%2B%202%5Cright%29%20%5Cleft%28x%20%2B%2010%5Cright%29%20%7D%7B%20x%20%2B%202%20%7D%20%3D%20x%20%2B%2010%20%20" alt="LaTeX: \frac{ x^{2} + 12 x + 36 - 16 }{ x + 2 } = \frac{ x^{2} + 12 x + 20 }{ x + 2 } = \frac{ \left(x + 2\right) \left(x + 10\right) }{ x + 2 } = x + 10 " data-equation-content=" \frac{ x^{2} + 12 x + 36 - 16 }{ x + 2 } = \frac{ x^{2} + 12 x + 20 }{ x + 2 } = \frac{ \left(x + 2\right) \left(x + 10\right) }{ x + 2 } = x + 10 " /> The reduced function is continuous at <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX: \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> and by the evaluation theorem <img class="equation_image" title=" \displaystyle \lim_{x \to -2 }x + 10 = 8 " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-2%20%7Dx%20%2B%2010%20%3D%208%20" alt="LaTeX: \displaystyle \lim_{x \to -2 }x + 10 = 8 " data-equation-content=" \displaystyle \lim_{x \to -2 }x + 10 = 8 " /> </p> </p>