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# For each function f, find the limit as x approaches c of the average rate of change of f from c to x. That is, find $\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$. $f(x)=\sqrt{2 x}, c=5$

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We would like to find the value of the average rate of change of $\color{#4257b2}\lim\limits_{x \to c}\dfrac{f(x)-f(c)}{x-c}$ if we know that $\color{#4257b2}f(x)=\sqrt{2x}$ and $\color{#4257b2}c=5$. First, we will find the average rate of change of $\color{#4257b2}f$ from $\color{#4257b2}c=5$ to $\color{#4257b2}x$ and then find the limit of this average rate of change.

$\dfrac{f(x)-f(c)}{x-c}=\dfrac{\sqrt{2x}-f(5)}{x-5}=\dfrac{\sqrt{2x}-\sqrt{2(5)}}{x-5}=\dfrac{\sqrt{2x}-\sqrt{10}}{x-5}$

Now we have the average rate of change, so the next step is to find the limit of this average as follows:

$\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}=\lim\limits_{x \to 5} \dfrac{\sqrt{2x}-\sqrt{10}}{x-5}$

Now we note that the value of the limit of the denominator at $\color{#4257b2}x=5$ will equal zero, so we need to use another strategy to find the value of the limit. We can multiply the numerator and denominator by $\color{#4257b2}\sqrt{2x}+\sqrt{10}$.

\begin{align*} \lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}&=\lim\limits_{x \to 5} \dfrac{\sqrt{2x}-\sqrt{10}}{x-5} \\ \\ &=\lim\limits_{x \to 5} \dfrac{\sqrt{2x}-\sqrt{10}}{x-5}\cdot \dfrac{\sqrt{2x}+\sqrt{10}}{\sqrt{2x}+\sqrt{10}} \\ \\ &=\lim\limits_{x \to 5} \dfrac{\left(\sqrt{2x}\right)^{2}-\left(\sqrt{10}\right)^{2}}{(x-5)\left(\sqrt{2x}+\sqrt{10}\right)} \\ \\ &=\lim\limits_{x \to 5} \dfrac{2x-10}{(x-5)\left(\sqrt{2x}+\sqrt{10}\right)} \\ \\ &=\lim\limits_{x \to 5} \dfrac{2\left(x-5\right)}{(x-5)\left(\sqrt{2x}+\sqrt{10}\right)} \\ \\ &=\lim\limits_{x \to 5} \dfrac{2\cancel{\left(x-5\right)}}{\cancel{(x-5)}\left(\sqrt{2x}+\sqrt{10}\right)}=\lim\limits_{x \to 5} \dfrac{2}{\sqrt{2x}+\sqrt{10}} \end{align*}

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