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

Building the fraction by multiplying by the conjugate in the numerator and denominator gives: \begin{equation*} \frac{ \sqrt{x + 3} - 2 }{ x - 1 }\left(\frac{ \sqrt{x + 3} + 2 }{ \sqrt{x + 3} + 2 } \right)=\frac{ x - 1 }{ \left(x - 1\right) \left(\sqrt{x + 3} + 2\right) } = \frac{1}{\sqrt{x + 3} + 2} \end{equation*}The reduced function is continuous at \(\displaystyle x = 1\) and by the evaluation theorem \(\displaystyle \lim_{x \to 1 }\frac{1}{\sqrt{x + 3} + 2} = \frac{1}{4}\)

Download \(\LaTeX\) 

\begin{question}Find the limit, if it exists $\lim_{x \to 1 }\frac{ \sqrt{x + 3} - 2 }{ x - 1 }$ 
    \soln{9cm}{Building the fraction by multiplying by the conjugate in the numerator and denominator gives: \begin{equation*} \frac{ \sqrt{x + 3} - 2 }{ x - 1 }\left(\frac{ \sqrt{x + 3} + 2 }{ \sqrt{x + 3} + 2 } \right)=\frac{ x - 1 }{ \left(x - 1\right) \left(\sqrt{x + 3} + 2\right) } = \frac{1}{\sqrt{x + 3} + 2} \end{equation*}The reduced function is continuous at $x = 1$ and by the evaluation theorem $\lim_{x \to 1 }\frac{1}{\sqrt{x + 3} + 2} = \frac{1}{4}$}

\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{ \sqrt{x + 3} - 2 }{ x - 1 } " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%201%20%7D%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%203%7D%20-%202%20%7D%7B%20x%20-%201%20%7D%20" alt="LaTeX:  \displaystyle \lim_{x \to 1 }\frac{ \sqrt{x + 3} - 2 }{ x - 1 } " data-equation-content=" \displaystyle \lim_{x \to 1 }\frac{ \sqrt{x + 3} - 2 }{ x - 1 } " />  </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 + 3} - 2 }{ x - 1 }\left(\frac{ \sqrt{x + 3} + 2 }{ \sqrt{x + 3} + 2 } \right)=\frac{ x - 1 }{ \left(x - 1\right) \left(\sqrt{x + 3} + 2\right) } = \frac{1}{\sqrt{x + 3} + 2}  " src="/equation_images/%20%20%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%203%7D%20-%202%20%7D%7B%20x%20-%201%20%7D%5Cleft%28%5Cfrac%7B%20%5Csqrt%7Bx%20%2B%203%7D%20%2B%202%20%7D%7B%20%5Csqrt%7Bx%20%2B%203%7D%20%2B%202%20%7D%20%5Cright%29%3D%5Cfrac%7B%20x%20-%201%20%7D%7B%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28%5Csqrt%7Bx%20%2B%203%7D%20%2B%202%5Cright%29%20%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%20%2B%203%7D%20%2B%202%7D%20%20" alt="LaTeX:   \frac{ \sqrt{x + 3} - 2 }{ x - 1 }\left(\frac{ \sqrt{x + 3} + 2 }{ \sqrt{x + 3} + 2 } \right)=\frac{ x - 1 }{ \left(x - 1\right) \left(\sqrt{x + 3} + 2\right) } = \frac{1}{\sqrt{x + 3} + 2}  " data-equation-content="  \frac{ \sqrt{x + 3} - 2 }{ x - 1 }\left(\frac{ \sqrt{x + 3} + 2 }{ \sqrt{x + 3} + 2 } \right)=\frac{ x - 1 }{ \left(x - 1\right) \left(\sqrt{x + 3} + 2\right) } = \frac{1}{\sqrt{x + 3} + 2}  " /> The reduced function is continuous at  <img class="equation_image" title=" \displaystyle x = 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201%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 }\frac{1}{\sqrt{x + 3} + 2} = \frac{1}{4} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%201%20%7D%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%20%2B%203%7D%20%2B%202%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20" alt="LaTeX:  \displaystyle \lim_{x \to 1 }\frac{1}{\sqrt{x + 3} + 2} = \frac{1}{4} " data-equation-content=" \displaystyle \lim_{x \to 1 }\frac{1}{\sqrt{x + 3} + 2} = \frac{1}{4} " /> </p> </p>