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Given epsilon = 1, find the largest value of delta graphically and algebraically
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\[\lim_{x \rightarrow 10} = \sqrt{19-x}\]
I assume you meant to write \(\lim\limits_{x\to10}\sqrt{19-x}=3\) ? You're looking for \(\delta\) such that \(|x-10|<\delta\) returns \(|\sqrt{19-x}-3|<1\). Solve this inequality for \(x\): \[\begin{align*} -1&<\sqrt{19-x}-3<1\\[1ex] 2&<\sqrt{19-x}<4\\[1ex] 4&<19-x<16\\[1ex] -15&<-x<-3\\[1ex] 3&<x<15 \end{align*}\]From here, you want to extract information about \(|x-10|\), which can be done by further solving explicitly for \(x-10\): \[3<x<15\implies -7<x-10<5\]which means the maximum distance in absolute value that \(x\) can be from \(10\) that guarantees \(|\sqrt{19-x}-3|<1\) must be \(\delta=5\).
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