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Calculus
Limits
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Find the limit, if it exists \(\displaystyle \lim_{x \to 4 }\frac{ \sqrt{x + 5} - 3 }{ x - 4 }\)

Building the fraction by multiplying by the conjugate in the numerator and denominator gives: \begin{equation*} \frac{ \sqrt{x + 5} - 3 }{ x - 4 }\left(\frac{ \sqrt{x + 5} + 3 }{ \sqrt{x + 5} + 3 } \right)=\frac{ x - 4 }{ \left(x - 4\right) \left(\sqrt{x + 5} + 3\right) } = \frac{1}{\sqrt{x + 5} + 3} \end{equation*}The reduced function is continuous at \(\displaystyle x = 4\) and by the evaluation theorem \(\displaystyle \lim_{x \to 4 }\frac{1}{\sqrt{x + 5} + 3} = \frac{1}{6}\)

Download \(\LaTeX\) 

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

\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 4 }\frac{ \sqrt{x + 5} - 3 }{ x - 4 } " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%204%20%7D%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%205%7D%20-%203%20%7D%7B%20x%20-%204%20%7D%20" alt="LaTeX:  \displaystyle \lim_{x \to 4 }\frac{ \sqrt{x + 5} - 3 }{ x - 4 } " data-equation-content=" \displaystyle \lim_{x \to 4 }\frac{ \sqrt{x + 5} - 3 }{ x - 4 } " />  </p> </p>
HTML for Canvas
<p> <p>Building the fraction by multiplying by the conjugate in the numerator and denominator gives:  <img class="equation_image" title="  \frac{ \sqrt{x + 5} - 3 }{ x - 4 }\left(\frac{ \sqrt{x + 5} + 3 }{ \sqrt{x + 5} + 3 } \right)=\frac{ x - 4 }{ \left(x - 4\right) \left(\sqrt{x + 5} + 3\right) } = \frac{1}{\sqrt{x + 5} + 3}  " src="/equation_images/%20%20%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%205%7D%20-%203%20%7D%7B%20x%20-%204%20%7D%5Cleft%28%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%205%7D%20%2B%203%20%7D%7B%20%5Csqrt%7Bx%20%2B%205%7D%20%2B%203%20%7D%20%5Cright%29%3D%5Cfrac%7B%20x%20-%204%20%7D%7B%20%5Cleft%28x%20-%204%5Cright%29%20%5Cleft%28%5Csqrt%7Bx%20%2B%205%7D%20%2B%203%5Cright%29%20%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%20%2B%205%7D%20%2B%203%7D%20%20" alt="LaTeX:   \frac{ \sqrt{x + 5} - 3 }{ x - 4 }\left(\frac{ \sqrt{x + 5} + 3 }{ \sqrt{x + 5} + 3 } \right)=\frac{ x - 4 }{ \left(x - 4\right) \left(\sqrt{x + 5} + 3\right) } = \frac{1}{\sqrt{x + 5} + 3}  " data-equation-content="  \frac{ \sqrt{x + 5} - 3 }{ x - 4 }\left(\frac{ \sqrt{x + 5} + 3 }{ \sqrt{x + 5} + 3 } \right)=\frac{ x - 4 }{ \left(x - 4\right) \left(\sqrt{x + 5} + 3\right) } = \frac{1}{\sqrt{x + 5} + 3}  " /> The reduced function is continuous at  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  and by the evaluation theorem  <img class="equation_image" title=" \displaystyle \lim_{x \to 4 }\frac{1}{\sqrt{x + 5} + 3} = \frac{1}{6} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%204%20%7D%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%20%2B%205%7D%20%2B%203%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20" alt="LaTeX:  \displaystyle \lim_{x \to 4 }\frac{1}{\sqrt{x + 5} + 3} = \frac{1}{6} " data-equation-content=" \displaystyle \lim_{x \to 4 }\frac{1}{\sqrt{x + 5} + 3} = \frac{1}{6} " /> </p> </p>