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I have no idea where to start with this problem. I have the graph drawn but do not know how to find delta.

If $5<x<5 + \delta$ then $\dfrac{x^2}{\sqrt{x-5}} > 100$

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Is the denominator of the fraction supposed to be $\sqrt{x}-5$ or $\sqrt{x-5}$? – Zev Chonoles Feb 3 '14 at 19:50
√x-5 square root of x-5, all of it! – Caitlin Feb 3 '14 at 19:57

For any $x>5$, we have that $\dfrac{\sqrt{x-5}}{100}$ is a positive number and we can safely multiply both sides of an inequality by it. Therefore, $$\frac{x^2}{\sqrt{x-5}}>100\iff \frac{x^2}{100}>\sqrt{x-5}.$$ (The $\iff$ arrow means "the left side is true if and only if the right side is true". In other words, it indicates that they are equivalent statements.)

Given positive $a$ and $b$, we have $a>b\iff a^2>b^2$, so that (with $a=\dfrac{x^2}{100}$ and $b=\sqrt{x-5}$), $$\frac{x^2}{\sqrt{x-5}}>100\iff \frac{x^2}{100}>\sqrt{x-5}\iff \frac{x^4}{10000}>x-5.$$ For any $x>5$, we have $$\frac{x^4}{10000}>\frac{5^4}{10000}=\frac{625}{10000}=0.0625.$$ Thus, as long as $x-5<0.0625$ - in other words, as long as $x<5+\delta$ where $\delta=0.0625$ - we get $$\frac{x^4}{10000}>0.0625>x-5,$$ and we know (by our earlier work) that this means $x$ satisfies $$\frac{x^2}{\sqrt{x-5}}>100.$$ Thus, for all $5<x<5+\delta$ where $\delta=0.0625$, we have $\dfrac{x^2}{\sqrt{x-5}}>100$.

(Of course, any smaller $\delta>0$ will also work.)

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