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Questions: Algebra BusinessCalculus

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Calculus
Limits
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Find the limit, if it exists \(\displaystyle \lim_{x \to -1 }\frac{ \left(x + 3\right)^{2} - 4 }{ x + 1 }\)

Expanding the numerator and simplifying gives: \begin{equation*} \frac{ x^{2} + 6 x + 9 - 4 }{ x + 1 } = \frac{ x^{2} + 6 x + 5 }{ x + 1 } = \frac{ \left(x + 1\right) \left(x + 5\right) }{ x + 1 } = x + 5 \end{equation*}The reduced function is continuous at \(\displaystyle x = -1\) and by the evaluation theorem \(\displaystyle \lim_{x \to -1 }x + 5 = 4\)

Download \(\LaTeX\) 

\begin{question}Find the limit, if it exists $\lim_{x \to -1 }\frac{ \left(x + 3\right)^{2} - 4 }{ x + 1 }$ 
    \soln{9cm}{Expanding the numerator and simplifying gives: \begin{equation*} \frac{ x^{2} + 6 x + 9 - 4 }{ x + 1 } = \frac{ x^{2} + 6 x + 5 }{ x + 1 } =  \frac{ \left(x + 1\right) \left(x + 5\right) }{ x + 1 } = x + 5 \end{equation*}The reduced function is continuous at $x = -1$ and by the evaluation theorem $\lim_{x \to -1 }x + 5 = 4$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the limit, if it exists  <img class="equation_image" title=" \displaystyle \lim_{x \to -1 }\frac{ \left(x + 3\right)^{2} - 4 }{ x + 1 } " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-1%20%7D%5Cfrac%7B%20%5Cleft%28x%20%2B%203%5Cright%29%5E%7B2%7D%20-%204%20%7D%7B%20x%20%2B%201%20%7D%20" alt="LaTeX:  \displaystyle \lim_{x \to -1 }\frac{ \left(x + 3\right)^{2} - 4 }{ x + 1 } " data-equation-content=" \displaystyle \lim_{x \to -1 }\frac{ \left(x + 3\right)^{2} - 4 }{ x + 1 } " />  </p> </p>
HTML for Canvas
<p> <p>Expanding the numerator and simplifying gives:  <img class="equation_image" title="  \frac{ x^{2} + 6 x + 9 - 4 }{ x + 1 } = \frac{ x^{2} + 6 x + 5 }{ x + 1 } =  \frac{ \left(x + 1\right) \left(x + 5\right) }{ x + 1 } = x + 5  " src="/equation_images/%20%20%5Cfrac%7B%20x%5E%7B2%7D%20%2B%206%20x%20%2B%209%20-%204%20%7D%7B%20x%20%2B%201%20%7D%20%3D%20%5Cfrac%7B%20x%5E%7B2%7D%20%2B%206%20x%20%2B%205%20%7D%7B%20x%20%2B%201%20%7D%20%3D%20%20%5Cfrac%7B%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%20%7D%7B%20x%20%2B%201%20%7D%20%3D%20x%20%2B%205%20%20" alt="LaTeX:   \frac{ x^{2} + 6 x + 9 - 4 }{ x + 1 } = \frac{ x^{2} + 6 x + 5 }{ x + 1 } =  \frac{ \left(x + 1\right) \left(x + 5\right) }{ x + 1 } = x + 5  " data-equation-content="  \frac{ x^{2} + 6 x + 9 - 4 }{ x + 1 } = \frac{ x^{2} + 6 x + 5 }{ x + 1 } =  \frac{ \left(x + 1\right) \left(x + 5\right) }{ x + 1 } = x + 5  " /> The reduced function is continuous at  <img class="equation_image" title=" \displaystyle x = -1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-1%20" alt="LaTeX:  \displaystyle x = -1 " data-equation-content=" \displaystyle x = -1 " />  and by the evaluation theorem  <img class="equation_image" title=" \displaystyle \lim_{x \to -1 }x + 5 = 4 " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-1%20%7Dx%20%2B%205%20%3D%204%20" alt="LaTeX:  \displaystyle \lim_{x \to -1 }x + 5 = 4 " data-equation-content=" \displaystyle \lim_{x \to -1 }x + 5 = 4 " /> </p> </p>